Infosys Specialist Programmer Coding Questions and Answers | Infosys Specialist Programmer Important Coding Questions | Infosys SP Coding Question with Answer | Infosys Specialist Programmer Coding Questions PDF Download
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Infosys Specialist Programmer Coding Questions
Company Name | Infosys |
Name of the Post | Specialist Programmer |
Category | Coding Questions and Answers |
Official website | infosys.com |
Infosys Specialist Programmer Coding Questions with Answers
Question 1:
Problem Statement :
A subarray of array A is a segment of contiguous elements in array A.
Given an array A of N elements, you can apply the following operations as many times as you like:
– Choosing a subarray [L, R] and subtracting 1 from each element in this subarray. The cost of this operation is X.
– Choosing an index i such that A[i] is positive, and setting A[i] = 0. The cost of this operation in Y.
Your task is to make all the elements equal to 0 and find the minimum cost to do so.
Input Format
- The first line contains an integer, N., denoting the number of elements in A.
- The next line contains an integer, X, denoting the cost of the first operation.
- The next line contains an integer. Y, denoting the cost of the second operation
- Each line i of the N subsequent lines (where 1 <=i<= N) contains an Integer describing Ai.
Constraints
- 1<=N<=10^5
- 1<=X<=10
- 1<=Y<=10^4
- 1<=A[i]<=10^8
Sample Input 1
1
1
10
1
Sample Output 1
1
Explanation:
N=1 X=1 Y=10 A=[1]. The optimal solution is to perform one operation of the first type on the subarray [1,N].
Sample Input 2
3
1
1
1
1
1
Sample Output 2
1
Explanation:
N=3 X=1 Y=1 A=[1,1,1] The optimal solution is to perform one operation of the first type on the subarray[1,N];
C++
#include <iostream> using namespace std; int main () { int arr[] = { 1, 1, 1 }; int X = 1; int Y = 1; int ans = 0; int arrSize = sizeof (arr) / sizeof (arr[0]); for (int i = 0; i < arrSize; i++) { arr[i]--; } ans = ans + X; for (int i = 0; i < arrSize; i++) { if (arr[i] != 0) { arr[i] = 0; ans = ans + Y; } } cout << ans; return 0; }
JAVA
n = 1 x = 1 y = 10 arr = [1] ans = 0 arrSize = len(arr) for i in range(arrSize): arr[i] = arr[i] - 1 ans = ans + x for i in range(arrSize): if arr[i] != 0: arr[i] = 0 ans = ans + y print(ans)
Problem Statement :
Wael is well-known for how much he loves the bitwise XOR operation, while kaito is well known for how much he loves to sum numbers, so their friend Resli decided to make up a problem that would enjoy both of them. Resil wrote down an array A of length N, an integer K and he defined a new function called Xor- sum as follows
- Xor-sum(x)=(x XOR A[1])+(x XOR A[2])+(x XOR A[3])+…………..+(x XOR A[N])
Can you find the integer x in the range [0,K] with the maximum Xor-sum (x) value?
Print only the value.
Input format
- The first line contains integer N denoting the number of elements in A.
- The next line contains an integer, k, denoting the maximum value of x.
- Each line i of the N subsequent lines(where 0<=i<=N) contains an integer describing Ai.
Constraints
- 1<=N<=10^5
- 0<=K<=10^9
- 0<=A[i]<=10^9
Sample Input 1
1
0
989898
Sample Output 1
989898
Explanation:
Xor_sum(0)=(0^989898)=989898
Sample Input 2
3
7
1
6
3
Sample Output 2
14
Explanation
Xor_sum(4)=(4^1)+(4^6)+(4^3)=14.
Sample Input 3
4
9
7
4
0
3
Sample Output 3
46
Explanation:
Xor_sum(8)=(8^7)+(8^4) +(8^0)+(8^3)=46.
C++
#include<bits/stdc++.h> using namespace std; unordered_map < int, int >L; int main () { int n, k, m; cin >> n >> k; vector < int >v (n); for (int i = 0; i < n; i++) { cin >> m; v[i] = m; int j = 0; while (m) { L[j] += (m & 1); m >>= 1; j++; } } int j = 0, K = k, ans = 0, ans2 = 0; while (K) { j++; K >>= 1; } for (int i = j; i > 0; i--) { if (L[i - 1] < n - L[i - 1]) ans != 1; ans <<= 1; } ans >>= 1; while (ans > k) { ans &= 0; ans <<= 1; k <<= 1; } for (auto i:v) ans2 += ans ^ i; cout << ans2; }
JAVA
import java.util.*; class Main { public static void main (String[]args) { Scanner sc = new Scanner (System.in); int n = sc.nextInt (); int k = sc.nextInt (); int arr[] = new int[n]; for (int i = 0; i < n; i++) arr[i] = sc.nextInt (); int res = 0, max = Integer.MIN_VALUE; for (int i = 0; i <= k; i++) { res = 0; for (int j = 0; j < n; j++) res = res + (i ^ arr[j]); max = Math.max (res, max); } System.out.println (max); } }
PHYTHON
def Xor_sum(x, arr): xorSum = sum(arr) for i in range(1, x): s = 0 for j in arr: s += i ^ j if s > xorSum: xorSum = s return xorSum n = 4 x = 9 arr = [7, 4, 0, 3] print(Xor_sum(x, arr))
Question 3:
Problem Statement :
Given an array A of N elements. You should choose a value B such that (B>=0), and then for each element in A set A[i]=A[i](+)B where is the bitwise XOR.
Print the minimum number of inversions in array A that you can achieve after choosing the value of B optimally and setting A[i] = A[i] (+) B. Since the answer might be large, print it modulo (10^9+7)
Input Format
- The first line contains an integer, N. denoting the number of elements in A
- Then the next line contains N elements, denoting the elements in A.
Input :
4
1 0 3 2
Output
1
C++
Find the smallest and largest number in an Array
Answer:
Sample Input
[3, 1, 56, 34, 12, 9, 98, 23, 4]C++
JAVA
PHYTHON
arr = [3, 1, 56, 34, 12, 9, 98, 23, 4] minVal = min(arr) maxVal = max(arr) print(f"Smallest Number: {minVal}") print(f"Largest Number: {maxVal}")
Output
Smallest Number: 1
Largest Number: 98
Question 5:
How do you rotate a matrix by 90 degrees? Write a program.
Answer:
Sample Input:
1 2 3
4 5 6
7 8 9
PHYTHON
JAVA
public static void main(String[] args) { // Input matrix
}
Output:
Rotated Matrix:
7 4 1
8 5 2
9 6 3
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