Ion Group Coding Questions and Answers
Ion Group Coding Questions With Solutions
Ion Group Coding Questions and Answers page will help you to prepare yourself to appear for recruitment drives for the designation of Software Engineer and Technical Analyst in Ion Group.
The page is mainly focused on the Coding Questions and Steps of the recruitment that are included in Online Assessment and Technical Interview of the company.
Sample Ion Group Coding Questions with Solutions
Question 1: Nearest smaller Tower
Given an array representing the heights of towers, the task is to find, for each tower, the index of the nearest tower that is shorter than it.
The search for a shorter tower can be performed by looking to the left and right sides of each tower.
The following rules apply:
If there are two or more smaller towers at the same distance from the current tower, choose the tower with the smallest height.
If two towers have the same height, choose the one with the smaller index.
Example 1:
Input : Array: [1, 3, 2]
Output : Indexes: [-1, 0, 0]
Explanation:
For the tower at index 0, there is no tower shorter than it, so the output is -1.
For the tower at index 1 (height 3), there are two towers (heights 1 and 2) at the same distance. Following the rules, we choose the tower with the smallest height, which is at index 0.
For the tower at index 2 (height 2), the only tower shorter than it is at index 0.
Therefore, the final output is the array of indexes: [-1, 0, 0].
Example 2:
Input : Array : [4, 8, 3, 5, 3]
Output : Indexes: [2, 2, -1, 2, -1]
Explanation:
For the tower at index 0 (height 4), the nearest tower shorter than it is at index 2.
For the tower at index 1 (height 8), there are two towers (heights 4 and 3) at the same distance.
Following the rules, we choose the tower at index 2.
For the tower at index 2 (height 3), there is no tower shorter than it.
For the tower at index 3 (height 5), there are two towers (heights 3 and 3) at the same distance.
Following the rules, we choose the tower at index 2 because it has a smaller index.
For the tower at index 4 (height 3), there is no tower shorter than it.
Therefore, the final output is the array of indexes: [2, 2, -1, 2, -1].
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#include<bits/stdc++.h>
class TowerFinder
{
public:
std::vector findNearestShorterTower(std::vector &towerHeights)
{
int n = towerHeights.size();
std::stack leftStack, rightStack;
std::vector result(n, -1);
for (int i = 0; i < n; i++) { while (!leftStack.empty() && towerHeights[leftStack.top()] >= towerHeights[i])
{
leftStack.pop();
}
if (!leftStack.empty())
{
result[i] = leftStack.top();
}
leftStack.push(i);
}
for (int i = n - 1; i >= 0; i--)
{
while (!rightStack.empty() && towerHeights[rightStack.top()] >= towerHeights[i])
{
rightStack.pop();
}
if (!rightStack.empty())
{
if (result[i] != -1)
{
if (std::abs(result[i] - i) == std::abs(rightStack.top() - i))
{
if (towerHeights[result[i]] > towerHeights[rightStack.top()])
result[i] = rightStack.top();
}
else if (std::abs(result[i] - i) > std::abs(rightStack.top() - i))
result[i] = rightStack.top();
}
else
result[i] = rightStack.top();
}
rightStack.push(i);
}
return result;
}
};
int main()
{
std::vector towerHeights = {4,8,3,5,3};
TowerFinder towerFinder;
std::vector nearestShorterTowers = towerFinder.findNearestShorterTower(towerHeights);
// Print the result
for (int i = 0; i < nearestShorterTowers.size(); i++)
{
std::cout << nearestShorterTowers[i] << " ";
}
std::cout << std::endl;
return 0;
}
class TowerFinder:
def find_nearest_shorter_tower(self, arr):
suff = []
n = len(arr)
find_suff = [0] * n # building suffix
for i in range(n - 1, -1, -1):
if len(suff) == 0:
find_suff[i] = -1
suff.append([arr[i], i])
else:
while suff:
if suff[-1][0] < arr[i]:
find_suff[i] = suff[-1][1]
break
suff.pop()
if len(suff) == 0:
find_suff[i] = -1
suff.append([arr[i], i])
pre = []
find_pre = [0] * n # building prefix
for i in range(n):
if len(pre) == 0:
find_pre[i] = -1
pre.append([arr[i], i])
else:
while pre:
if pre[-1][0] < arr[i]: find_pre[i] = pre[-1][1] break pre.pop() if len(pre) == 0: find_pre[i] = -1 pre.append([arr[i], i]) new = [0] * n # comparing both for i in range(n): if find_suff[i] == -1: new[i] = find_pre[i] continue if find_pre[i] == -1: new[i] = find_suff[i] continue if abs(find_suff[i] - i) == abs(find_pre[i] - i): if arr[find_suff[i]] >= arr[find_pre[i]]:
new[i] = find_pre[i]
else:
new[i] = find_suff[i]
elif abs(find_suff[i] - i) > abs(find_pre[i] - i):
new[i] = find_pre[i]
else:
new[i] = find_suff[i]
return new
# Test the code
tower_heights = [1, 3, 2]
tower_finder = TowerFinder()
nearest_shorter_towers = tower_finder.find_nearest_shorter_tower(tower_heights)
# Print the result
print(nearest_shorter_towers)
import java.util.*;
class Main {
public static int[] findNearestShorterTower(int[] towerHeights) {
int n = towerHeights.length;
Stack leftStack = new Stack<>();
Stack rightStack = new Stack<>();
int[] result = new int[n];
Arrays.fill(result, -1);
for (int i = 0; i < n; i++) { while (!leftStack.empty() && towerHeights[leftStack.peek()] >= towerHeights[i]) {
leftStack.pop();
}
if (!leftStack.empty()) {
result[i] = leftStack.peek();
}
leftStack.push(i);
}
for (int i = n - 1; i >= 0; i--) {
while (!rightStack.empty() && towerHeights[rightStack.peek()] >= towerHeights[i]) {
rightStack.pop();
}
if (!rightStack.empty()) {
if (result[i] != -1) {
if (Math.abs(result[i] - i) == Math.abs(rightStack.peek() - i)) {
if (towerHeights[result[i]] > towerHeights[rightStack.peek()]) {
result[i] = rightStack.peek();
}
} else if (Math.abs(result[i] - i) > Math.abs(rightStack.peek() - i)) {
result[i] = rightStack.peek();
}
} else {
result[i] = rightStack.peek();
}
}
rightStack.push(i);
}
return result;
}
public static void main(String[] args) {
int[] towerHeights = {4,8,3,5,3};
int[] nearestShorterTowers = findNearestShorterTower(towerHeights);
// Print the result
for (int i = 0; i < nearestShorterTowers.length; i++) {
System.out.print(nearestShorterTowers[i] + " ");
}
System.out.println();
}
}
Question 2:
Problem Statement – Abhijeet is one of those students who tries to get his own money by part time jobs in various places to fill up the expenses for buying books. He is not placed in one place, so what he does, he tries to allocate how much the book he needs will cost, and then work to earn that much money only. He works and then buys the book respectively. Sometimes he gets more money than he needs so the money is saved for the next book. Sometimes he doesn’t. In that time, if he has stored money from previous books, he can afford it, otherwise he needs money from his parents.
Now His parents go to work and he can’t contact them amid a day. You are his friend, and you have to find how much money minimum he can borrow from his parents so that he can buy all the books.
He can Buy the book in any order.
Function Description:
Complete the function with the following parameters:
| Name | Type | Description |
| N | Integer | How many Books he has to buy that day. |
| EarnArray[ ] | Integer array | Array of his earnings for the ith book |
| CostArray[ ] | Integer array | Array of the actual cost of the ith book. |
Constraints:
- 1 <= N <= 10^3
- 1 <= EarnArray[i] <= 10^3
- 1 <= CostArray[i] <= 10^3
Input Format:
- First line contains N.
- Second N lines contain The ith earning for the ith book.
- After that N lines contain The cost of the ith book.
Output Format: The minimum money he needs to cover the total expense.
Sample Input 1:
3
[3 4 2]
[5 3 4]
Sample Output 1:
3
Explanation:
At first he buys the 2nd book, which costs 3 rupees, so he saves 1 rupee. Then he buys the 1st book, that takes 2 rupees more. So he spends his stored 1 rupee and hence he needs 1 rupee more. Then he buys the last book.
#include<bits/stdc++.h>
using namespace std;
int main() {
int n, ans = 0, sum = 0;
cin >> n;
vector< int > arr1(n), arr2(n);
for (int i = 0; i < n; i++) { cin >> arr2[i];
}
for (int i = 0; i < n; i++) { cin >> arr1[i];
}
for (int i = 0; i < n; i++) {
arr2[i] -= arr1[i];
}
sort(arr2.begin(), arr2.end(), greater< int >());
for (int i = 0; i < n; i++) {
sum += arr2[i];
if (sum < 0) {
ans += abs(sum);
sum = 0;
}
}
cout << ans << endl;
return 0;
}
n = int(input())
ans = 0
su = 0
arr2 = list(map(int, input().split()))
arr1 = list(map(int, input().split()))
for i in range(n):
arr2[i] -= arr1[i]
arr2.sort(reverse=True)
for i in range(n):
su += arr2[i]
if su < 0:
ans += abs(su)
su = 0
print(ans)
import java.util.*;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner (System.in);
int n=sc.nextInt();
int arr1[]= new int[n];
int arr2[]= new int[n];
for(int i=0; i<n; i++)
arr2[i]=sc.nextInt();
for(int i=0; i<n; i++)
arr1[i]=sc.nextInt();
for(int i=0; i<n; i++) arr2[i] -= arr1[i]; Arrays.sort(arr2); int sum=0, ans=0; for(int i=n-1; i>=0; i--)
{
sum+=arr2[i];
if(sum<0)
{
sum = -sum;
ans= ans +sum ;
sum=0;
}
}
System.out.println(ans);
}
}
Question 3: Queries for Count
The task is to determine the number of elements within a specified range in an unsorted array. Given an array of size n, the goal is to count the elements that fall between two given values, i and j, inclusively.
Examples:
Input:
Array: [1, 3, 3, 9, 10, 4]
Range 1: i = 1, j = 4
Range 2: i = 9, j = 12
Output:
For Range 1: 4
For Range 2: 2
Explanation:
In the first query, the numbers within the range 1 to 4 are 1, 3, 3, and 4.
In the second query, the numbers within the range 9 to 12 are 9 and 10.
#include<bits/stdc++.h>
using namespace std;
int findLowerIndex(int arr[], int n, int x) {
int low = 0, high = n - 1;
while (low <= high) { int mid = (low + high) / 2; if (arr[mid] >= x)
high = mid - 1;
else
low = mid + 1;
}
return low;
}
int findUpperIndex(int arr[], int n, int y) {
int low = 0, high = n - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] <= y)
low = mid + 1;
else
high = mid - 1;
}
return high;
}
int countElementsInRange(int arr[], int n, int x, int y) {
int count = 0;
count = findUpperIndex(arr, n, y) - findLowerIndex(arr, n, x) + 1;
return count;
}
int main() {
int arr[] = {1, 4, 4, 9, 10, 3};
int n = sizeof(arr) / sizeof(arr[0]);
sort(arr, arr + n);
int lowerRange = 1, upperRange = 4;
cout << countElementsInRange(arr, n, lowerRange, upperRange) << endl;
lowerRange = 9;
upperRange = 12;
cout << countElementsInRange(arr, n, lowerRange, upperRange) << endl;
return 0;
}
def find_lower_index(arr, x):
low, high = 0, len(arr) - 1
while low <= high: mid = (low + high) // 2 if arr[mid] >= x:
high = mid - 1
else:
low = mid + 1
return low
def find_upper_index(arr, y):
low, high = 0, len(arr) - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] <= y:
low = mid + 1
else:
high = mid - 1
return high
def count_elements_in_range(arr, x, y):
arr.sort()
lower_index = find_lower_index(arr, x)
upper_index = find_upper_index(arr, y)
count = upper_index - lower_index + 1
return count
arr = [1, 4, 4, 9, 10, 3]
lower_range = 1
upper_range = 4
print(count_elements_in_range(arr, lower_range, upper_range))
lower_range = 9
upper_range = 12
print(count_elements_in_range(arr, lower_range, upper_range))
import java.util.Arrays;
class Main {
static int findLowerIndex(int[] arr, int n, int x) {
int low = 0, high = n - 1;
while (low <= high) { int mid = (low + high) / 2; if (arr[mid] >= x)
high = mid - 1;
else
low = mid + 1;
}
return low;
}
static int findUpperIndex(int[] arr, int n, int y) {
int low = 0, high = n - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (arr[mid] <= y)
low = mid + 1;
else
high = mid - 1;
}
return high;
}
static int countElementsInRange(int[] arr, int n, int x, int y) {
int count = 0;
count = findUpperIndex(arr, n, y) - findLowerIndex(arr, n, x) + 1;
return count;
}
public static void main(String[] args) {
int[] arr = {1, 4, 4, 9, 10, 3};
int n = arr.length;
Arrays.sort(arr);
int lowerRange = 1, upperRange = 4;
System.out.println(countElementsInRange(arr, n, lowerRange, upperRange));
lowerRange = 9;
upperRange = 12;
System.out.println(countElementsInRange(arr, n, lowerRange, upperRange));
}
}
Question 4 : Copycat
Problem Statement :
Ashish was copying from Rahit in the exam. So, Rahit told him to change the answers a little bit so that the examiner cannot find the fraud. But silly Ashish in the way started to change all the answers that were needed. He shuffled the letters in each word in a way where the maximum number of letters were misplaced.
For a given word, find the maximum difference that Ashish can generate between his answer and Rahit’s answer.
Suppose Rahit wrote “car” for an answer, Ashish can write “acr” with difference 2, or “arc” with differnece 3.
Note That: The letters are all in lowercase.
Input Format:
First line containing an integer n, number of words.
Then, n numbers of lines as the query words.
Output:
N number of lines with an integer each denoting possible maximum difference.
Sample Input:
4
abababa
bbj
kj
kk
Sample Output:
6
2
2
0
#include<bits/stdc++.h>
using namespace std;
string s, s1;
int n;
int func() {
if (n <= 1) return 0;
int ans = 0, c = 0;
sort(s.begin(), s.end());
for (int i = 0; i < n; i++)
if (s1[i] != s[i]) c++;
ans = max(ans, c);
c = 0;
while (next_permutation(s.begin(), s.end())) {
for (int i = 0; i < n; i++) if (s1[i] != s[i]) c++; ans = max(ans, c); c = 0; } return ans; } int main() { int t; cin >> t;
while (t--) {
cin >> s;
s1 = s;
n = s.length();
cout << func() << endl;
}
return 0;
}
import java.util.*;
class Main {
public static String swapString(String s, int i, int j) {
char[] b = s.toCharArray();
char temp;
temp = b[i];
b[i] = b[j];
b[j] = temp;
return String.valueOf(b);
}
public static void generatePermutation(String s, int start, int end, HashSet < String > set) {
if (start == end - 1)
set.add(s + " ");
else {
for (int i = start; i < end; i++) {
s = swapString(s, start, i);
generatePermutation(s, start + 1, end, set);
s = swapString(s, start, i);
}
}
}
public static int maxDiff(String str, String s) {
int c = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == str.charAt(i));
else
c++;
}
return c;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String[] a = new String[n];
for (int i = 0; i < n; i++) {
a[i] = sc.next();
}
for (int i = 0; i < n; i++) {
HashSet < String > set = new HashSet < String > ();
String s = a[i];
generatePermutation(s, 0, s.length(), set);
int max = 0;
int k = 0;
for (String str: set) {
k = maxDiff(str, s);
max = Math.max(max, k);
}
System.out.println(max);
}
}
}
from itertools import permutations
def solve(x, n):
ans = 0
for i in permutations(x):
c = 0
for j in range(n):
if x[j] != i[j]:
c += 1
ans = max(ans, c)
return ans
t = int(input())
for i in range(t):
s = input()
print(solve(s, len(s)))
Question 5 : Death Note
Problem Statement :
Ryuk, the Shinigami (God of death) had allowed Light Yagami, a school student, to kill as many people as he can by using a death note. But writing the names barely will allow other people to watch them. So he encrypts the names using digits, where a means 1, b means 2 and so on upto z is 26. Now if he gives numbers, there is a communication error because a number string can be decrypted by the death note in various ways and eventually killing them all. If everyone in the world has a unique name, for a given number, how many people can die?
NOTE THAT: There is every possible name in the world with the 26 letters, and capital or small letters is not a problem.
Input format:
A number stream denoting the first name’s encrypted version
Output Format:
Number of people dying by this.
Constraints:
1<=stream length<=10^8
Sample Input: 1267
Sample Output: 3
Output Specification:Two people of the name azg and lfg die.
#include <bits/stdc++.h>
using namespace std;
string s;
int ans;
void func(int i, int n) {
if (i == n - 1 || i == n) {
ans++;
return;
}
if (s[i] == '1')
func(i + 2, n);
else if (s[i] == '2' && s[i + 1] < '7') func(i + 2, n); func(i + 1, n); } int main() { ans = 0; cin >> s;
func(0, s.length());
cout << ans;
}
import java.util.*;
class Main {
static String s;
static int ans;
public static void func(int i, int n) {
if (i == n - 1 || i == n) {
ans++;
return;
}
if (s.charAt(i) == '1')
func(i + 2, n);
else if (s.charAt(i) == '2' && s.charAt(i + 1) < '7')
func(i + 2, n);
func(i + 1, n);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
s = sc.next();
func(0, s.length());
System.out.println(ans);
}
}
def solve(s, n):
if n == 0 or n == 1:
return 1
ans = 0
if s[n - 1] > "0":
ans = solve(s, n - 1)
if s[n - 2] == "1" or (s[n - 2] == "2" and s[n - 1] < "7"):
ans += solve(s, n - 2)
return ans
s = input()
print(solve(s, len(s)))
Question 6 : Choco and chocolate
Problem Statement :
Choco, a chocolate lover, has N amount of money with him. He wants to buy as much chocolate as possible. So, he goes to a chocolate shop “Bandyman ”. Mike, the owner of “Bandyman ” has different types of chocolate in his store (represented by a character) placed in a row.
Mike, give an offer to Choco that he can buy a selected type of chocolate for free and need to pay for the other types of chocolates and Choco can only buy consecutive chocolates.
Now, you need to write a code to find the maximum amount of chocolates Choco can get by selecting the chocolates optimally.
Input format :
1st line contains 2 space separated integers A and B denoting the number of chocolates and the amount of money Choco has.
The 2nd line contains A chocolates represented by a string. All chocolates represented by lowercase alphabets.
The 3rd line represents 26 space separated integers representing the cost to buy the chocolates.
[First integer represents the cost of the chocolate of type ‘a’, 2nd integer represents the cost of the chocolates of type ‘b’ and so on]
Output format :
Print the maximum number of chocolates Choco can buy.
Constraints :
1<=A<=10^5
1<=B<=10^9
1<=cost of chocolate<=10^9
Sample input 1 :
6 10
aabcda
5 4 4 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Sample output 1 :
4
Explanation :
Choco can select the chocolate of type ‘a’ for free and start buying from index 0 and if he buys “aabc” then he has to pay less (0+0+4+4=8) than the total money he has.
This is the maximum number of chocolates he can get in this case.
#include<bits/stdc++.h>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
string s, s1;
cin >> s;
cin.ignore();
getline(cin, s1);
istringstream ss(s1);
vector < int > a;
vector < char > a1;
while (ss) {
string w;
ss >> w;
if (w == "") break;
a.push_back(stoi(w));
}
map < char, int > mm;
for (auto i: s)
if (mm[i] == 0) {
mm[i]++;
a1.push_back(i);
}
int t = 0;
for (auto i: a1) {
vector < int > a2;
for (auto i1: s) {
if (i1 == i) a2.push_back(0);
else a2.push_back(a[i1 - 97]);
}
int l = 0, su = 0;
for (int i2 = 0; i2 < n; i2++) { if (l == i2) su += a2[i2]; else su += a2[i2]; if (su > m) {
while (su > m && l <= i2) { su -= a2[l]; l++; } } if (i2 - l + 1 > t) t = i2 - l + 1;
}
}
cout << t;
}
import java.util.*;
class Main {
public static int solve(int n, int amnt, String s, int[] price) {
int[] freq = new int[26];
char[] ch = s.toCharArray();
int temp = 0, ans = 0, maxFreq = 0, st = 0;
for (int i = 0; i < n; i++) { int indx = ch[i] - 'a'; freq[indx] += price[indx]; temp += price[indx]; maxFreq = Math.max(maxFreq, freq[indx]); if (temp - maxFreq > amnt) {
while (temp > amnt) {
boolean b = false;
int tempIndx = ch[st] - 'a';
if (maxFreq == freq[tempIndx]) {
b = true;
}
temp -= price[tempIndx];
freq[tempIndx] -= price[tempIndx];
st++;
if (b) {
maxFreq = 0;
for (int j = 0; j < 26; j++) {
maxFreq = Math.max(freq[j], maxFreq);
}
}
if (temp - maxFreq < amnt) {
break;
}
}
}
ans = Math.max(ans, i - st + 1);
}
return ans;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int amnt = sc.nextInt();
String s = sc.next();
int[] price = new int[26];
for (int i = 0; i < n; i++) {
price[i] = sc.nextInt();
}
System.out.println(solve(n, amnt, s, price));
}
}
n, m = map(int, input().split())
s = input().strip()
a = list(map(int, input().split()))
s1 = set()
for i in s:
if i not in s1:
s1.add(i)
s1 = list(s1)
t = 0
for i in s1:
a1 = []
for i1 in s:
if i1 == i:
a1.append(0)
else:
a1.append(a[ord(i1) - 97])
l = 0
su = 0
for i2 in range(n):
if l == i2:
su = a1[i2]
else:
su += a1[i2]
if su > m:
while su > m and l <= i2: su -= a1[l] l += 1 if (i2 - l + 1) > t:
t = i2 - l + 1
print(t)
QUESTION 7: Formatting large Products
Problem Statement: Rohan is weak in mathematics.He is giving mathematics Olympiad , but he got stuck in one of the question .Help rohan to solve the question.In Question there are two positive integer A and B. You have to find the product of all integer between A and B which is represented in the form C=D*10^E , where C is the product of numbers , D and E are non-negative integers and the last digit of D is non-zero.
Function Description
- Complete the function formatProducts in the editor below, formatProduct must return a string that represents C in the above described form.
- Function has the following parameters
- A: an integer
- B: an integer
Constraints
- A will between 1 and 1,000,000 . Inclusive.
- B will be between A and 1,000,000. Inclusive.
Sample Input 0
- 1
- 5
Sample Output 0
- 12 * 10^1
Explanation
- 1*2*3*4*5=120 = 12 * 10^1
Sample Input 1
- 3
- 10
Sample Output 1
- 18144 * 10^2
Explanation
- 3*4*….*10=1814400 =18144 * 10^2
import java.util.*;
public class Main
{
public static String formatProducts (int a, int b)
{
int result = 1;
for (int i = a; i <= b; i++)
result = result * i;
int temp = result;
int power = 0;
while ((result % 10) == 0)
{
power = power + 1;
result = result / 10;
}
return result + " * 10^" + power;
}
public static void main (String[]args)
{
Scanner sc = new Scanner (System.in);
int a = sc.nextInt ();
int b = sc.nextInt ();
System.out.println (formatProducts (a, b));
}
}
def formatProducts(L,H):
a=1
for i in range(L,H+1):
a=a*i
c=0
while(True):
if a%10!=0:
break
else:
c+=1
a=a//10
return (str(a)+" * 10^"+str(c))
L=int(input())
H=int(input())
print(formatProducts(L,H))
Question 8: Coloured Zenga
Problem Statement :
Rahul is playing a game, wherein he has multiple coloured wooden blocks, stacked one above the other, his task is to remove all the wooden blocks from the stack, without letting it fall and in the minimum number of steps. He can remove one block of a colour at a time, but he can remove multiple blocks of the same colour together. Determine the minimum number of steps in which he can perform this task.
For example, if you remove [red,red] from (white,red,red,white), the resulting array is [white,white].
Note- there are only two colour blocks – red and white
Function description :
Complete the minMoves function in the provided editor. It contains the following parameters:
Parameters:
| Name | Type | Description |
|---|---|---|
| N | Integer | No. of Wooden blocks |
| Array[ ] | Integer Array | Array of Blocks. |
Input format :
The first line contains an integer n denoting the number of blocks. Each n line denotes the colour of the wooden block .
Constraints :
1<=n<=700
0<=a[i]<=1
Sample input 1 :
4
red
white
white
red
Sample Output 2 :
2
Explanation :
Remove [white,white] first The array will be [red,red] The remaining numbers can be removed in one strap .
Sample Input 1:
4
white
red
white
red
Sample Output 1:
3
Sample Explanation:
0
The steps are [white,red,white,red]->[red,white,red]->[red,red]->[]. Therefore the answer is 3.
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector < string > arr(n);
for (int i = 0; i < n; i++) cin >> arr[i];
int a = 0, b = 0;
bool start = false;
for (int i = 0; i < n; i++) {
if (arr[i] == "white") {
if (!start) {
start = true;
}
} else {
if (start) {
a++;
start = false;
}
}
}
if (start) a++;
start = false;
for (int i = 0; i < n; i++) {
if (arr[i] == "red") {
if (!start) {
start = true;
}
} else {
if (start) {
b++;
start = false;
}
}
if (start) b++;
}
cout << min(a + 1, b + 1) << endl;
return 0;
}
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String arr[] = new String[n];
for (int i = 0; i < n; i++)
arr[i] = sc.next();
int a = 0, b = 0;
boolean flag = false;
for (int i = 0; i < n; i++) {
if (arr[i].equals("white")) {
if (!flag)
flag = true;
} else {
if (flag) {
a++;
flag = false;
}
}
}
if (flag)
a++;
flag = false;
for (int i = 0; i < n; i++) {
if (arr[i].equals("red")) {
if (!flag)
flag = true;
} else {
if (flag) {
b++;
flag = false;
}
}
if (flag)
b++;
}
System.out.println(Math.min(a + 1, b + 1));
}
}
n = int(input())
arr = []
for i in range(n):
arr.append(input())
a, b = 0, 0
start = False
for i in range(n):
if arr[i] == "white":
if not start:
start = True
else:
if start:
a += 1
start = False
if start:
a += 1
start = False
for i in range(n):
if arr[i] == "red":
if not start:
start = True
else:
if start:
b += 1
start = False
if start:
b += 1
print(min(a + 1, b + 1))
Question 9:
Sahil watches TV all day and gets bored. He started playing this dumb game of identifying minimum number of inputs needed to reach a channel. As his cousin, you have to help him, but you live far from his house. So you decide to write a code that will ask Sahil for some inputs and give outputs respectively.
Here are the problems you need to keep in mind,
- There are 13 buttons on his remote: 10 buttons for the numbers (0-9) to form integers denoting respective channel index, “Up channel” button and “ Down channel” button for going i +1th channel and i-1th channel from i respectively, and a “Last viewed” button to see what’s the last channel before it.
- The number buttons allow you to jump directly to a specific channel (Ex: to go to channel 172 by typing 1,7,2).
- If the channel which you are in is ith and that is the max channel index possible, by Up channel, you will reach the first channel possible. Same goes for the down channel button. You can go to the highest channel possible if you go down from the lowest channel possible.
Sahil can get from one channel to the next in one of the two ways.
Sahil’s parents have set some parental control on some channels on Aniruth’s television. The “Up Channel “ and “Down buttons” buttons skip these channels as these channels are not viewable.
Given a list of channels to view, the lowest channel, the highest channel, and a list of blocked channels, your program should return the minimum number of clicks necessary to get through all the shows that Anirudh would like to match.
Input Format
- First line is the lowest Channel
- Second-line is the highest Channel
- Followed by a number of blocked channels B,
- and the next B lines contain the actual blocked channels.
- Followed by the number of Channels to view V, and the next V lines contain the actual channels to view.
Constraints
- The lowest channel on the television will be greater than 0. and less than or equal to 10,000.
- The highest channel on the television will be greater than or equal to the lowest channel. and less than or equal to 10.000.
- The list of channels that are blocked on Anirudh’s television. All the channels in this list will be valid channels (greater than or equal to lowest channel, less than or equal 1 to highest channel). Duplicates may be Ignored. The blocked list can be a maximum of 40 channels.
- The sequence that Sahil must view contains between 1 and 50 elements. inclusive. All channels in this sequence are not in the blocked list and are between lowest channel and highest channel. Inclusive.
Examples
- Sample Input 0:
1
20
2
18
19
5
15
14
17
1
17
- Sample output 0:
7
#include<bits/stdc++.h>
using namespace std;
unordered_map < int, int > m;
int l, u;
util(int a, int b) {
if (a == b) return 0;
if (m[a]) return util(a + 1, b);
return 1 + util(a + 1, b);
}
int func(int b, int prev) {
if (b < prev) return min(util(prev, u) + util(l, b) + 1, util(b, prev)); else return min(util(prev, b), util(l, b) + util(prev, u) + 1); } int main() { int flag = 0, ans = 0, prev, prev2; cin >> l >> u;
int bn, b;
cin >> bn;
while (bn--) {
cin >> b;
m[b]++;
}
cin >> bn;
while (bn--) {
cin >> b;
if (b > 9 && flag == 0) {
ans += 2;
flag++;
prev = b;
} else if (flag == 0) {
ans += 1;
flag++;
prev = b;
} else if (prev2 == b) {
prev2 = prev;
prev = b;
ans++;
} else {
ans += min(b > 9 ? 2 : 1, func(prev, b));
prev2 = prev;
prev = b;
}
}
cout << ans;
}
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int l = sc.nextInt();
int h = sc.nextInt();
int b = sc.nextInt();
List < Integer > bl = new ArrayList < > ();
for (int i = 0; i < b; i++) {
bl.add(sc.nextInt());
}
int v = sc.nextInt();
List < Integer > vl = new ArrayList < > ();
for (int i = 0; i < v; i++) {
vl.add(sc.nextInt());
}
Set < Integer > sl = new HashSet < > ();
int res = 0;
for (Integer i: vl) {
if (bl.contains(i))
continue;
sl.add(i);
}
for (Integer i: sl) {
String s = i + "";
res += s.length();
}
System.out.println(res);
}
}
Question 10: Square Free Numbers Problem
Problem Description
In the theory of numbers, square free numbers have a special place. A square free number is one that is not divisible by a perfect square (other than 1). Thus 72 is divisible by 36 (a perfect square), and is not a square free number, but 70 has factors 1, 2, 5, 7, 10, 14, 35 and 70. As none of these are perfect squares (other than 1), 70 is a square free number.
For some algorithms, it is important to find out the square free numbers that divide a number. Note that 1 is not considered a square free number.
In this problem, you are asked to write a program to find the number of square free numbers that divide a given number.
Input:-
- The only line of the input is a single integer N which is divisible by no prime number larger than 19.
Output:-
- One line containing an integer that gives the number of square free numbers (not including 1).
Constraints:-
- N < 10^9
Complexity:-
- Simple
Time Limit:-
- 1
Examples
Example 1
- Input:-
20 - Output:-
3
Explanation
N=20
If we list the numbers that divide 20, they are
1, 2, 4, 5, 10, 20
1 is not a square free number, 4 is a perfect square, and 20 is divisible by 4, a perfect square. 2 and 5, being prime, are square free, and 10 is divisible by 1,2,5 and 10, none of which are perfect squares. Hence the square free numbers that divide 20 are 2, 5, 10. Hence the result is 3.
Example 2
- Input:-
72 - Output:-
3
Explanation
N=72. The numbers that divide 72 are
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
1 is not considered square free. 4, 9 and 36 are perfect squares, and 8,12,18,24 and 72 are divisible by one of the. Hence only 2, 3 and 6 are square free. (It is easily seen that none of them are divisible by a perfect square). The result is 3.
#include<string.h>
#include<stdio.h>
int main()
{
long int n;
double sqnum;
long int i,j=0,flag,chksqr,temp[10000];
int count=0,k;
printf("Enter the number:");
scanf("%ld",&n);
for(i=2;i<=n/2;i++)
{
if(n%i==0)
{
count++;
sqnum=sqrt(i);
chksqr=sqnum;
if(chksqr==sqnum)
{
count--;
temp[j]=i;
j++;
}
else
{
for(k=0;k<j;k++) { if(i>temp[k] && j!=0)
{
if(i%temp[k]==0)
{
count--;
k=j+1;
}
}
else
break;
}
}
}
}
printf("%d",count);
return 0;
}
Output Enter the number : 20 3
#include<bits/stdc++.h>
using namespace std;
int main()
{
long int n;
double sqnum;
long int i,j=0,flag,chksqr,temp[10000];
int count=0,k;
cout<<"Enter the number:";
cin>>n;
for(i=2;i<=n/2;i++)
{
if(n%i==0)
{
count++;
sqnum=sqrt(i);
chksqr=sqnum;
if(chksqr==sqnum)
{
count--;
temp[j]=i;
j++;
}
else
{
for(k=0;k<j;k++) { if(i>temp[k] && j!=0)
{
if(i%temp[k]==0)
{
count--;
k=j+1;
}
}
else
break;
}
}
}
}
cout<<count;
return 0;
}
Output Enter the number : 20 3
FAQs on Ion Group Coding Questions with Solutions
Question 1: How many rounds are there in the Ion Group Recruitment Exam?
Total Rounds in the Ion Group Recruitment exam is:-
- Online Assessment
- Technical Interview
- HR Interview
Question 2: What questions are included in Technical Section of Online Assessment of Ion Group?
DSA and Core Computer Science Based MCQs are Included in Ion Group’s Online Assessments. And apart from that DSA Based Coding Questions were asked in Technical Interview.
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