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Interview Preparation
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Off Campus
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 |
Wipro coding questions range from easy to difficult level. Out of the 2 questions asked in Wipro coding test, one is easy and the other is slightly difficult. You need to answer a minimum of 1 question, to clear the cut off. But in certain cases, if a lot of students have cleared all test cases for both the questions, then the cutoff might vary accordingly.
Wipro coding test questions pattern and Facts –
| Wipro Coding Test Questions and Answers Type | Probability of Being Asked | Number of Question(Max) | Time Alloted |
|---|---|---|---|
| Wipro Coding Pattern Question | 50% | 1 | within 60 mins |
| Wipro Coding Round Searching and Sorting Questions | 20% | 1 | within 60 mins |
| Functional Questions Like HCF and GCD | 20% | 0 – 1 | within 45 mins |
| Others | 10% | 1 | within 45 mins |
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
int sockMerchant(int n, int ar_count, int* ar) { int i, j, count; int pair = 0; int unique=0; int new_ar[n]; for(i=0; i<ar_count; i++) { count = 0; for(j=0;j<=i;j++) { if(ar[i]==ar[j]) { count++; } } if(count==1) { new_ar[unique] = ar[i]; unique++; } } for(i=0; i<unique; i++) { count = 0; for(j=0; j<ar_count; j++) { if(new_ar[i]==ar[j]) { count++; } } pair = pair + (count/2); } return pair; }
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> #include <string.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 == 0) { 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; }
def countingValley(steps, path): level = valley = 0 for i in path: if i == 'U': level += 1 elif i == 'D': if level == 0: 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]); } }
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; }
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); } }
#Python3 program to print #given matrix in spiral form def spiralPrint (m, n, a): 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 and l < n): #Print the first row from #the remaining rows for i in range (l, n): print (a[k][i], end = " ") k += 1 #Print the last column from #the remaining columns for i in range (k, m): print (a[i][n - 1], end = " ") n -= 1 #Print the last row from #the remaining rows if (k < m): for i in range (n - 1, (l - 1), -1): print (a[m - 1][i], end = " ") m -= 1 #Print the first column from #the remaining columns if (l < n): for i in range (m - 1, k - 1, -1): print (a[i][l], end = " ") l += 1 #Driver Code a =[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12],[13, 14, 15, 16, 17, 18]] R = 3; C = 6 spiralPrint (R, C, a)
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 <stdio.h> #include <math.h> // 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; }
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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int main()
{
int n,d;
cin >> n >> d;
int arr[n];
for(int i=0;i> arr[i];
for(int i=(d%n) ;i<n+(d%n) ; i++)
cout << arr[i%n];
}
for question 3
#include
using namespace std;
int main()
{
int n,d;
cin >> n >> d;
int arr[n];
for(int i=0;i> arr[i];
for(int i=(d%n);i<n+(d%n);i++)
cout << arr[i%n] << " ";
}
In which language is the automata question given?will we get the option to select the programming language in that?
We provide the coding questions in C and C++
what are the programming languages that will be given for the Wipro turbo challenge coding round?
Here are the languages thar are choosable in the Exam:
C
C++
Java
Python
coding in wipro is tough or moderate
Hi Ananya,
Coding is moderate in the Wipro Exam.
i have ordered both coding and verbal sections for wipro, amount for both materials is debited but i m nt able to access the material,plz check.
Hi shilpa,
You can find the steps to access the Paid Materials from here -https://prepinsta.com/steps-to-access-paid-material/
Thanks Prepinsta, i gave exam yesterday around 2 questions got repeated from this page and around 40-50% of the questions were similar or repeated from your online classes :).
Is there any course for interview preparation also
Hi Saurabh, we are really glad that your exam went well.
And for Interview Preparation, you may just go through our Interview Experiences Dashboards – https://prepinsta.com/interview-experience/Wipro/
Wipro Coding Questions
Yes, Python is allowed as an coding language.
You can check detailed syllabus for Wipro here – https://prepinsta.com/wipro-syllabus/