Wipro Coding Questions is a very important section in Wipro off-campus or on-campus drives. In this round there are 2 problem statement, for which the students have to provide code for, in any of the mentioned languages they like to. The below are some of the practice questions that will help you in your preparation for Wipro Coding Section.
| Coding Questions | Wipro |
| Time | 60 mins |
| Number of Questions | 2 |
| Coding Difficulty | Medium |
| Adaptive Test | No |
| Coding Questions | Wipro |
| Time | 60 mins |
| Number of Questions | 2 |
| Coding Difficulty | Medium |
| Adaptive Test | No |
Wipro coding test questions pattern and Facts –
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Find the latest syllabus for Wipro here on this page.
Problem Statement
Alex works at a clothing store. There is a large pile of socks that must be paired by color for sale. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are.
For example, there are n=7 socks with colors ar = {1,2,1,2,1,3,2}. There is one pair of color 1 and one of color 2. There are three odd socks left, one of each color. The number of pairs is 2.
Function Description
Complete the sockMerchant function in the editor below. It must return an integer representing the number of matching pairs of socks that are available.
sockMerchant has the following parameter(s):
n: the number of socks in the pile
ar: the colors of each sock
Input Format
The first line contains an integer n, the number of socks represented in ar.
The second line contains n space-separated integers describing the colors ar[i] of the socks in the pile.
Constraints
1 <= n <= 100
1 <= ar[i] <= 100 & 0 <= i < n
Output Format
Return the total number of matching pairs of socks that Alex can sell.
Sample Input
9
10 20 20 10 10 30 50 10 20
Sample Output
3
Explanation
Alex can match 3 pairs of socks i.e 10-10, 10-10, 20-20
while the left out socks are 50, 60, 20
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int sockMerchant(int n, int arr[])
{
int freq[101]={0};
int ans = 0,i;
for(i=0;i<n;i++)
freq[arr[i]]++;
for(i = 0; i <= 100; i++)
{
ans = ans+ freq[i]/2;
}
return ans;
}
int main ()
{
int n;
scanf("%d",&n);
int arr[101],i;
for (i = 0; i < n; i++)
{
scanf("%d",&arr[i]);
}
int ans=sockMerchant(n,arr);
printf("%d\n",ans);
return 0;
}
#include<bits/stdc++.h>
using namespace std;
int sockMerchant(int n, int arr[])
{
int freq[101]={0};
int ans = 0;
for(int i=0;i<n;i++)
{
int value=arr[i];
freq[value]++;
}
for(int i = 0; i <= 100; i++)
{
ans = ans+ freq[i]/2;
}
return ans;
}
int main ()
{
int n;
cin >> n;
int arr[n]={0};
for (int i = 0; i < n; i++)
{
cin>>arr[i];
}
int res=sockMerchant(n,arr);
cout<<res<<endl;
return 0;
}
import java.util.*;
class Main
{
public static int sockMerchant(int n, int arr[])
{
int freq[]=new int[101];
for(int i=0;i<n;i++)
{
freq[arr[i]]++;
}
int ans=0;
for(int i=0;i<=100;i++)
ans=ans+freq[i]/2;
return ans;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++)
arr[i]=sc.nextInt();
int ans=sockMerchant(n,arr);
System.out.println(ans);
}
}
def sockMerchant(n, ar): pairs = 0 set_ar = set(ar) for i in set_ar: count = ar.count(i) pairs+=count//2 return pairs n = int(input()) ar = list(map(int, input().split())) print(sockMerchant(n, ar))
Problem Statement
Gary is an avid hiker. He tracks his hikes meticulously, paying close attention to small details like topography. During his last hike, he took exactly n steps. For every step he took, he noted if it was an uphill or a downhill step. Gary’s hikes start and end at sea level. We define the following terms:
Given Gary’s sequence of up and down steps during his last hike, find and print the number of valleys he walked through.
Input Format
The first line contains an integer, , denoting the number of steps in Gary’s hike.
The second line contains a single string of characters. Each character belongs to {U, D} (where U indicates a step up and D indicates a step down), and the i(th) cin the string describes Gary’s i(th) step during the hike.
Constraints
Output Format
Print a single integer denoting the number of valleys Gary walked through during his hike.
Sample Input
8
UDDDUDUU
Sample Output
1
Explanation
If we represent _ as sea level, a step up as / , and a step down as \ , Gary’s hike can be drawn as:
_/\ _
\ /
\/\/
It’s clear that there is only one valley there, so we print on a new line.
#include<stdio.h> int countingValley(long int steps, char *path) { long int i, level = 0, valley = 0; for(i=0; i<steps; i++) { if(path[i] == 'U') { level++; } else if(path[i] == 'D') { if(level == 1) { valley++; } level--; } } return valley; } int main() { long int steps; scanf("%li",&steps); char path[steps]; int i, result; for(i=0; i<steps; i++) { scanf("%c",&path[i]); } result = countingValley(steps, path); printf("%d",result); return 0; }
#include<bits/stdc++.h>
using namespace std;
int countingValley(long int steps, char *path)
{
long int i, level = 0, valley = 0;
for(i=0; i<steps; i++)
{
if(path[i] == 'U')
{
level++;
}
else if(path[i] == 'D')
{
if(level == 1)
{
valley++;
}
level--;
}
}
return valley;
}
int main()
{
long int steps;
cin>>steps;
char path[steps];
int i, result;
for(i=0; i<steps; i++)
{
cin>>path[i];
}
result = countingValley(steps, path);
cout<<result;
return 0;
}
import java.util.*;
class Main
{
public static int countingValley(int n, String path)
{
int level=0,valley=0;
for(int i=0;i<n;i++)
{
if(path.charAt(i)=='U')
level++;
else if(path.charAt(i)=='D')
{
if(level==1)
valley++;
level--;
}
}
return valley;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
String str=sc.next();
int result=countingValley(n,str);
System.out.println(result);
}
}
def countingValley(steps, path): level = valley = 0 for i in path: if i == 'U': level += 1 elif i == 'D': if level == 1: valley += 1 level -= 1 return valley steps = int(input()) path = input() print(countingValley(steps, path))
Problem Statement
A left rotation operation on an array shifts each of the array’s elements unit to the left. For example, if 2 left rotations are performed on array [1, 2, 3, 4, 5], then the array would become [3, 4, 5, 1, 2].
Given an array of integers and a number, , perform left rotations on the array. Return the updated array to be printed as a single line of space-separated integers.
Function Description
Complete the function rotLeft in the editor below. It should return the resulting array of integers.
rotLeft has the following parameter(s):
Input Format
The first line contains two space-separated integers and , the size of and the number of left rotations you must perform.
The second line contains space-separated integers a[i].
Constraints
Output Format
Print a single line of space-separated integers denoting the final state of the array after performing d left rotations.
Sample Input
5 4
1 2 3 4 5
Sample Output
5 1 2 3 4
Explanation
When we perform d=4 left rotations, the array undergoes the following sequence of changes:
[1,2,3,4,5] → [2,3,4,5,1] → [3,4,5,1,2] → [4,5,1,2,3] → [5,1,2,3,4]
Test Case : 1
Input (stdin)
Expected Output
Test Case : 2
Input (stdin)
Expected Output
#include<stdio.h> int rotLeft(int arr[], int n, int d) { int i, j; int first; for(i=0; i<d; i++) { first = arr[0]; for(j=0; j<n-1; j++) { arr[j] = arr[j+1]; } arr[j] = first; } return *arr; } int main() { int n, d, i; scanf("%d",&n); scanf("%d",&d); int list[n]; for(i=0; i<n; i++) { scanf("%d",&list[i]); } rotLeft(list, n, d); for(i=0; i<n; i++) { printf("%d ",list[i]); } }
#include<bits/stdc++.h>
using namespace std;
int rotLeft(int arr[], int n, int d)
{
int i, j;
int first;
for(i=0; i<d; i++)
{
first = arr[0];
for(j=0; j<n-1; j++)
{
arr[j] = arr[j+1];
}
arr[j] = first;
}
return *arr;
}
int main()
{
int n, d, i;
cin>>n;
cin>>d;
int list[n];
for(i=0; i<n; i++)
{
cin>>list[i];
}
rotLeft(list, n, d);
for(i=0; i<n; i++)
{
cout<<list[i]<<" ";
}
}
import java.util.*;
class Main
{
public static void rotateLeft(int a[],int n, int d)
{
int first,i,j;
for(i=0;i<d;i++)
{
first=a[0];
for(j=0;j<n-1;j++)
a[j]=a[j+1];
a[j]=first;
}
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int d=sc.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
a[i]=sc.nextInt();
rotateLeft(a,n,d);
for(int i=0;i<n;i++)
System.out.print(a[i]+" ");
}
}
def rotateLeft(n,d,arr): for i in range(d): arr = rotatearr(n,arr) return arr def rotatearr(n, arr): first = arr[0] for i in range(n-1): arr[i] = arr[i+1] arr[n-1] = first return arr , d = map(int, input().split()) arr = list(map(int, input().split())) for i in rotateLeft(n,d,arr): print(i, end=" ")
Ques: C Program to check if two given matrices are identical
#include<stdio.h> #define N 4 // This function returns 1 if A[][] and B[][] are identical // otherwise returns 0 int areSame (int A[][N], int B[][N]) { int i, j; for (i = 0; i < N; i++) for (j = 0; j < N; j++) if (A[i][j] != B[i][j]) return 0; return 1; } int main () { int A[N][N] = { {1, 1, 1, 1}, {2, 2, 2, 2}, {3, 3, 3, 3}, {4, 4, 4, 4} }; int B[N][N] = { {1, 1, 1, 1}, {2, 2, 2, 2}, {3, 3, 3, 3}, {4, 4, 4, 4} }; if (areSame (A, B)) printf ("Matrices are identical "); else printf ("Matrices are not identical"); return 0; }
#include<bits/stdc++.h>
using namespace std;
#define N 4
// This function returns 1 if A[][] and B[][] are identical
// otherwise returns 0
int areSame (int A[][N], int B[][N])
{
int i, j;
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
if (A[i][j] != B[i][j])
return 0;
return 1;
}
int main ()
{
int A[N][N] = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}
};
int B[N][N] = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}
};
if (areSame (A, B))
cout<<"Matrices are identical";
else
cout<<"Matrices are not identical";
return 0;
}import java.util.*;
class Main
{
static int size=4;
public static boolean areSame(int A[][], int B[][])
{
int i,j;
for(i=0;i<size;i++)
{
for(j=0;j<size;j++)
if(A[i][j]!=B[i][j])
return false;
}
return true;
}
public static void main(String[] args)
{
int A[][]={{1,1,1,1},{2,2,2,2},{3,3,3,3,},{4,4,4,4}};
int B[][]={{1,1,1,1},{2,2,2,2},{3,3,3,3,},{4,4,4,4}};
if(areSame(A,B))
{
System.out.println("Matrices are identical");
}
else
System.out.println("Matrices are not identical");
}
}a = [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]] b = [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]] if a==b: prin("Metrices are identical") else: pint("Metrices are not identical")
Ques: Given a 2D array, print it in spiral form. See the following examples.
NOTE:- Please comment down the code in other languages as well below .
Input:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output:
1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10
Input:
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
Output:
1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11
#include<stdio.h> #define R 3 #define C 6 void spiralPrint (int m, int n, int a[R][C]) { int i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ while (k < m && l < n) { /* Print the first row from the remaining rows */ for (i = l; i < n; ++i) { printf ("%d ", a[k][i]); } k++; /* Print the last column from the remaining columns */ for (i = k; i < m; ++i) { printf ("%d ", a[i][n - 1]); } n--; /* Print the last row from the remaining rows */ if (k < m) { for (i = n - 1; i >= l; --i) { printf ("%d ", a[m - 1][i]); } m--; } /* Print the first column from the remaining columns */ if (l < n) { for (i = m - 1; i >= k; --i) { printf ("%d ", a[i][l]); } l++; } } } /* Driver program to test above functions */ int main () { int a[R][C] = { {1, 2, 3, 4, 5, 6}, {7, 8, 9, 10, 11, 12}, {13, 14, 15, 16, 17, 18} }; spiralPrint (R, C, a); return 0; }
#include <bits/stdc++.h> using namespace std; #define R 3 #define C 6 void spiralPrint(int m, int n, int a[R][C]) { int i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ while (k < m && l < n) { /* Print the first row from the remaining rows */ for (i = l; i < n; ++i) { cout << a[k][i] << " "; } k++; /* Print the last column from the remaining columns */ for (i = k; i < m; ++i) { cout << a[i][n - 1] << " "; } n--; /* Print the last row from the remaining rows */ if (k < m) { for (i = n - 1; i >= l; --i) { cout << a[m - 1][i] << " "; } m--; } /* Print the first column from the remaining columns */ if (l < n) { for (i = m - 1; i >= k; --i) { cout << a[i][l] << " "; } l++; } } } /* Driver program to test above functions */ int main() { int a[R][C] = { { 1, 2, 3, 4, 5, 6 }, { 7, 8, 9, 10, 11, 12 }, { 13, 14, 15, 16, 17, 18 } }; spiralPrint(R, C, a); return 0; }
// Java program to print a given matrix in spiral form import java.io.*; class Main { // Function print matrix in spiral form static void spiralPrint (int m, int n, int a[][]) { int i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ while (k < m && l < n) { // Print the first row from the remaining rows for (i = l; i < n; ++i) { System.out.print (a[k][i] + " "); } k++; // Print the last column from the remaining columns for (i = k; i < m; ++i) { System.out.print (a[i][n - 1] + " "); } n--; // Print the last row from the remaining rows */ if (k < m) { for (i = n - 1; i >= l; --i) { System.out.print (a[m - 1][i] + " "); } m--; } // Print the first column from the remaining columns */ if (l < n) { for (i = m - 1; i >= k; --i) { System.out.print (a[i][l] + " "); } l++; } } } // driver program public static void main (String[]args) { int R = 3; int C = 6; int a[][] = { {1, 2, 3, 4, 5, 6}, {7, 8, 9, 10, 11, 12}, {13, 14, 15, 16, 17, 18} }; spiralPrint (R, C, a); } }
def spiralOrder(arr):ans=[]while arr:ans+=arr.pop(0)arr= (list(zip(*arr)))[::-1]return ansarr=[[1, 2, 3, 4, 5, 6],[7, 8, 9, 10, 11, 12],[13, 14, 15, 16, 17, 18]]print(spiralOrder(arr))
Ques: A Pythagorean triplet is a set of three integers a, b and c such that a2 + b2 = c2. Given a limit, generate all Pythagorean Triples with values smaller than given limit.
Input : limit = 20
Output : 3 4 5
8 6 10
5 12 13
15 8 17
12 16 20A Simple Solution is to generate these triplets smaller than given limit using three nested loop. For every triplet, check if Pythagorean condition is true, if true, then print the triplet. Time complexity of this solution is O(limit3) where ‘limit’ is given limit.
An Efficient Solution can print all triplets in O(k) time where k is number of triplets printed. The idea is to use square sum relation of Pythagorean triplet, i.e., addition of squares of a and b is equal to square of c, we can write these number in terms of m and n such that,
a = m2 - n2
b = 2 * m * n
c = m2 + n2
because,
a2 = m4 + n4 – 2 * m2 * n2
b2 = 4 * m2 * n2
c2 = m4 + n4 + 2* m2 * n2
We can see that a2 + b2 = c2, so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets
// A C program to generate pythagorean triplets // smaller than a given limit #include #include // Function to generate pythagorean triplets // smaller than limit void pythagoreanTriplets (int limit) { // triplet: a^2 + b^2 = c^2 int a, b, c = 0; // loop from 2 to max_limitit int m = 2; // Limiting c would limit all a, b and c while (c < limit) { // now loop on j from 1 to i-1 for (int n = 1; n < m; ++n) { // Evaluate and print triplets using // the relation between a, b and c a = m * m - n * n; b = 2 * m * n; c = m * m + n * n; if (c > limit) break; printf ("%d %d %d \n", a, b, c); } m++; } } // Driver program int main () { int limit = 20; pythagoreanTriplets (limit); return 0; }
#include<bits/stdc++.h>
using namespace std;
// Function to generate pythagorean triplets
// smaller than limit
void pythagoreanTriplets (int limit)
{
// triplet: a^2 + b^2 = c^2
int a, b, c = 0;
// loop from 2 to max_limitit
int m = 2;
// Limiting c would limit all a, b and c
while (c < limit)
{
// now loop on j from 1 to i-1
for (int n = 1; n < m; ++n)
{
// Evaluate and print triplets using
// the relation between a, b and c
a = m * m - n * n;
b = 2 * m * n;
c = m * m + n * n;
if (c > limit)
break;
cout<<a<<" "<<b<<" "<<c<<endl;
}
m++;
}
}
// Driver program
int main ()
{
int limit = 20;
pythagoreanTriplets (limit);
return 0;
}
import java.util.*;
class Main
{
public static void pythagoreanTriplets(int limit)
{
int a,b,c=0;
int m=2;
while(c<limit)
{
for(int n=1;n<m;++n)
{
a=m*m-n*n;
b=2*m*n;
c=m*m+n*n;
if(c>limit)
break;
System.out.println(a+" "+b+" "+c);
}
m++;
}
}
public static void main(String[] args)
{
int limit=20;
pythagoreanTriplets(limit);
}
}def pythagoreanTriplets(limit): a = b = c = 0 m = 2 while True: for n in range(1,m+1): a = m*m - n*n b = 2*m*n c = m*m + n*n if c > limit: break if a==0 or b==0 or c==0: break print(a,b,c) m=m+1 limit = int(input()) pythagoreanTriplets(limit)
Ans. There are generally one easy question which will mostly be a pattern based program and the other question will be non pattern question like GCD of two numbers etc.
Ans. Wipro NLTH test is also conducted by Wipro but the level of difficutly is high, thus you can study from our Wipro NLTH Coding Questions Page.
Ans. Wipro Coding Questions are of moderate difficulty but you need to score atleast 70%ile in this section to get to the next round necessarily.
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#QUESTION 3
#python code
n,d=list(map(int,input().split()))
l=list(map(int,input().split()))
for i in range(d):
x=l[0]
l.remove(l[0])
l=l+[x]
k=”
for i in l:
k=k+’ ‘+str(i)
k=(k).strip(‘ ‘)
print(k)
question 3 sollution:-
def de(a,k):
v=a[k:]
d=[]
s=””
for i in a:
if i not in v:
d.append(i)
for i in d:
v.append(i)
for i in v:
s=s+str(i)+” ”
return s.strip()
a=[41, 73, 89, 7 ,10, 1, 59, 58, 84, 77, 77, 97, 58, 1, 86, 58, 26, 10, 86 ,51]
k=10
print(de(a,k))