Infosys Specialist Programmer (SP) Coding Questions and Answers, Get Infosys SP Coding Questions

Infosys Specialist Programmer Coding Questions and Answers | Infosys Specialist Programmer Important Coding Questions | Infosys SP Coding Question with Answer | Infosys SP Coding Questions 

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Infosys Specialist Programmer Coding Questions 2024

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

public class Main
{
  public static void main (String[]args)
  {

    int[] arr = { 1, 1, 1 };

    int X = 1;
    int Y = 1;
    int ans = 0;

    for (int i = 0; i < arr.length; i++)
      {

	arr[i]--;

      }
    ans = ans + X;

    for (int i = 0; i < arr.length; i++)
      {
	if (arr[i] != 0)
	  {
	    arr[i] = 0;
	    ans = ans + Y;
	  }
      }
    System.out.println (ans);
  }
}

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)

#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++

#include <bits/stdc++.h>
#define sz(x) ((int)(x).size())
#define inf mod

using namespace std;

const int maxn = (int) 5e6 + 100;
int n, t[2][maxn], id = 1;
int dp[2][30];
vector < int >g[maxn];

void add (int x, int pos)
{
  int v = 0;
  for (int i = 29; i >= 0; i--)
    {
      int bit = ((x >> i) & 1);
      if (!t[bit][v])
	t[bit][v] = id++;
      v = t[bit][v];
      g[v].push_back (pos);
    }
}

void go (int v, int b = 29)
{
  int l = t[0][v], r = t[1][v];
  if (l)
    go (l, b - 1);
  if (r)
    go (r, b - 1);
  if (!l || !r)
    return;
  int res = 0;
  int ptr = 0;
for (auto x:g[l])
    {
      while (ptr < sz (g[r]) && g[r][ptr] < x)
	ptr++;
      res += ptr;
    }
  dp[0][b] += res;
  dp[1][b] += sz (g[l]) * 1ll * sz (g[r]) - res;
}


int main ()
{
  cin >> n;
  for (int i = 1; i <= n; i++)
    {
      int x;
      cin >> x;
      add (x, i);
    }
  go (0);
  int inv = 0;
  int res = 0;
  for (int i = 0; i <= 29; i++)
    {
      inv += min (dp[0][i], dp[1][i]);
      if (dp[1][i] < dp[0][i])
	res += (1 << i);
    }
  cout << inv;
}

Find the smallest and largest number in an Array

Answer: 

Sample Input

C++

JAVA 

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}")

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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