Infosys Off Campus Coding Question and Answers 2025, Get Infosys Interview Coding Questions

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Infosys Off Campus Coding Question and Answers 2025: The Coding Questions and Answers is very important in the IT Sector and Private Job Sector. Now days candidates are mostly interested in the off-Campus drive and private jobs. Eagerly prepare for the Coding Interview, Assessment Test, and attend the many samples online test for Private jobs. Following of the hiring, Infosys issued the Off Campus Drive for eligible and capable candidates. The Infosys Off Campus Drive selection process coding test, technical test etc., The Coding Interview round will be asked many questions through sample of C, C++, Java, Python Programming Languages. Thoroughly prepare for the Infosys Off campus Coding Interview. Here we uploaded major common Infosys Off campus Coding Questions and Infosys Coding Interview Questions. Applicants must prepare Infosys Off Campus Coding Questions and Answers.

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Infosys Off Campus Interview Coding Questions 2024

Infosys Coding Interview and Answers 

Question 1: While playing an RPG game, you were assigned to complete one of the hardest quests in this game. There are n monsters you’ll need to defeat in this quest. Each monster i is described with two integer numbers – poweri and bonusi. To defeat this monster, you’ll need at least poweri experience points. If you try fighting this monster without having enough experience points, you lose immediately. You will also gain bonusi experience points if you defeat this monster. You can defeat monsters in any order.

The quest turned out to be very hard – you try to defeat the monsters but keep losing repeatedly. Your friend told you that this quest is impossible to complete. Knowing that, you’re interested, what is the maximum possible number of monsters you can defeat?

Input:

• The first line contains an integer, n, denoting the number of monsters. The next line contains an integer, e, denoting your initial experience.
• Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, poweri, which represents power of the corresponding monster.
• Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, bonusi, which represents bonus for defeating the corresponding monster.

JAVA

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

class Main
{
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int exp = s.nextInt();

        int monst[] = new int[n];
        int bonus[] = new int[n];

        for (int i = 0; i < n; i++) {
            monst[i] = s.nextInt();
        }
        for (int i = 0; i < n; i++) {
            bonus[i] = s.nextInt();
        }

        class Monster {
            private final int power, bonus;
            public Monster(int power, int bonus){
                this.power = power;
                this.bonus = bonus;
            }
        }

        Monster[] monsters = new Monster[n];

        for(int i = 0; i < n; i++)
            monsters[i] = new Monster(monst[i], bonus[i]);

        Arrays.sort(monsters, Comparator.comparingInt(m -> m.power));

        int count = 0;

        for(Monster m: monsters){
            if(exp < m.power) break;
            exp += m.bonus;
            ++count;
        }
        System.out.println(count);
    }
}

PHYTHON

n = int(input())
lev = int(input())
p = []
b = []
a = []
ans = 0
for i in range(n):
   p.append(int(input()))
for j in range(n):
   b.append(int(input()))
for k in range(n):
   a.append([p[k], b[k]])
a.sort()
for z in a:
   if z[0] > lev:
       break
   lev += z[1]
   ans += 1
print(ans)

C++

#include<iostream>
#include<map>

Question 2: You have an array A of N integers A1 A2 .. An. Find the longest increasing subsequence Ai1 Ai2 .. Ak
(1 <= k <= N) that satisfies the following condition:
For every adjacent pair of numbers of the chosen subsequence Ai[x] and Ai[x+1] (1 < x < k), the expression( Ai[x] & Ai[x+1] ) * 2 < ( Ai[x] | Ai[x+1] ) is true

Note: ‘&’ is the bitwise AND operation, ‘ | ‘ is the bit-wise OR operation

Input:

1. The first line contains an integer, N, denoting the number of elements in A.
2. Each line i of the N subsequent lines (where 0 ≤ i < N) contains an integer describing Ai.

Sample cases:

JAVA 

class Main
{
    public static int LIS(int[] arr, int i, int n, int prev)
    {
        if (i == n) {
            return 0;
        }
        int excl = LIS(arr, i + 1, n, prev);

        int incl = 0;
        if (arr[i] > prev) {
            incl = 1 + LIS(arr, i + 1, n, arr[i]);
        }

        return Integer.max(incl, excl);
    }

    public static void main(String[] args)
    {
        int[] arr = { 15, 6, 5, 12, 1 };

        System.out.print("The length of the LIS is "
                        + LIS(arr, 0, arr.length, Integer.MIN_VALUE));
    }
}

PHYTHON

def sub(arr, i, n, prev=0):
   if i == n:
       return 0
   a = sub(arr, i + 1, n, prev)
   b = 0
   if arr[i] > prev:
       b = 1 + sub(arr, i + 1, n, arr[i])
   return max(b, a)


n = int(input())
arr = []
for i in range(n):
   arr.append(int(input()))
print("Length of Bitwise subsequence will be", sub(arr, 0, len(arr)))

C++

#include<bits/stdc++.h>
using namespace std;
int sub(int arr[], int i, int n, int prev){
    if(i == n)
       return 0;
    int a = sub(arr, i + 1, n, prev);
    int b = 0;
    if (arr[i] > prev)
       b = 1 + sub(arr, i + 1, n, arr[i]);
   
    return max(b, a);
}

int main(){
    
    int n;
    cin>>n;
    
    int arr[n];
    
    for(int i=0; i< n; i++)
    cin>>arr[i];
    
    int res = sub(arr, 0, n, 0);
    cout<< res;
}

import java.util.*;
class Main
{
  static String str;
  static int cash, n, a, b;
  static void swapf ()
  {
    char s[] = str.toCharArray ();
    int i = 0;
    for (int a = 0; a < s.length; a++)
      if (s[a] == '1')
	{
	  i = a;
	  break;
	}
    int j = s.length - 1;
    while (j > i)
      {
	if (cash < a)
	  break;
	if (s[j] == '0')
	  {
	    if (s[i] == '0')
	      i++;
	    else
	      {
		char temp = s[i];
		s[i] = s[j];
		s[j] = temp;
		cash -= a;
		j--;
	      }
	  }
	else
	  j--;
      }
    str = new String (s);
  }
  static void flipf ()
  {
    char s[] = str.toCharArray ();
    int i = 0;

    for (int a = 0; a < s.length; a++)
      if (s[a] == '1')
	{
	  i = a;
	  break;
	}
    while (cash >= b)
      {
	if (i == s.length)
	  break;
	if (s[i] == '1')
	  {
	    s[i] = '0';
	    i++;
	    cash -= b;
	  }
      }
    str = new String (s);
  }

  public static void main (String[]args)
  {
    Scanner sc = new Scanner (System.in);
    n = sc.nextInt ();
    str = sc.next ();
    cash = sc.nextInt ();
    a = sc.nextInt ();
    b = sc.nextInt ();

    if (a < b)
      {
	swapf ();
	flipf ();
      }
    else
      {
	flipf ();
	swapf ();
      }
    System.out.println (Integer.parseInt (str, 2));
  }
}

PHYTHON 

# N = int(input())
# S = list(input())
# Cash = int(input())
# A = int(input())
# B = int(input())
N = 6
S = list("111011")
Cash = 7
A = 1
B = 3


def swap():
   global Cash
   Rs = S.copy()
   S[S.index('1')], S[''.join(S).rindex('0')] = S[''.join(S).rindex('0')], S[S.index('1')]
   if Rs == S:
       flip()
   else:
       Cash -= A


def flip():
   global Cash
   S[S.index('1')] = '0'
   Cash -= B


while Cash > A or Cash > B:
   if A < B and '0' in S:
       swap()
   else:
       flip()
print(int(''.join(S), 2))

#include<bits/stdc++.h>
using namespace std;
string s;
int n, cash, a, b;

void swapf ()
{
  int i;

  for (int a = 0; a < s.length (); a++) 
  if (s[a] == '1') 
  { 
      i = a; 
      break; 
      
  } 
  int j = s.length () - 1; 
  while (j > i)
    {
      if (cash < a)
	break;

      if (s[j] == '0')
	{
	  if (s[i] == '0')
	    i++;

	  else
	    {
	      swap (s[i], s[j]);
	      cash -= a;
	      j--;
	    }
	}
      else
	j--;
    }
}
void flipf ()
{
  int i;
  for (int a = 0; a < s.length (); a++) 
  if (s[a] == '1') 
  { 
      i = a; break; 
      
  } 
  while (cash >= b)
    {

      if (i == s.length ())
      break;

      if (s[i] == '1')
	{
	  s[i] = '0';
	  i++;
	  cash -= b;
	}
    }
}

int main ()
{
  cin >> n >> s >> cash >> a >> b;

  if (a < b)
    {
      swapf ();
      flipf ();
    }

  else
    {
      flipf ();
      swapf ();
    }

  cout << stoull (s, 0, 2);

}

Question 4: Khaled has an array A of N elements. It is guaranteed that N is even. He wants to choose at most N/2 elements from array A. It is not necessary to choose consecutive elements.  Khaled is interested in XOR of all the elements he chooses. Here, XOR denotes the bitwise XOR operation.

For example:

• If A=[2,4,6,8], then khaled can choose the subset [2,4,8] to achieve XOR=(2 XOR 4 XOR 8)=14.

Khaled wants to maximize the XOR of all the elements he chooses. Your task is to help khaled to find the max XOR of a subset that he can achieve by choosing at most N/2 elements?

   Input format:

• The first line contains an integer, N, denoting the number of elements in A.
• Each line i of the N subsequent lines(where 0<=i<=N) contains an integer describing Ai.

   Constraints 

• 1<=N<=120
• 1<=A[i]<=10^6

   Sample Input 1

2
1
2
   Sample Output 1
2

Explanation:

N=2,  A=[1,2] khaled can choose the subset[2]. The xor of the elements in the subset is 2. And the number of elements in the subset is 1 which is less than N/2.

Sample Input 2
4
1
2
4
7

Sample Output 2

7

Explanation:

N=4,  A=[1,2,4,7] Khaled can choose the subset [7]. The xor of the elements in the subset is 7, and the number of elements in the subset is 1 which is less than N/2.

JAVA

import java.util.*;
class Main
{
  public static void main (String[]args)
  {
    final int N = 120, M = 1 << 20;
    int dp[] = new int[M];
    char res[] = new char[M];

    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 ();

    for (int i = 1; i < M; i++)
        dp[i] = Integer.MAX_VALUE;

    for (int i = 0; i < n; ++i)
      {
	if (arr[i] == 0)
	  continue;
	for (int j = 0; j < M; ++j)
	  res[j] = 0;

	for (int k = 0; k < M; ++k) { if (res[k] == 1) continue; if (dp[k] > dp[k ^ arr[i]])
	      dp[k] = dp[k ^ arr[i]] + 1;
	      
	    else if (dp[k ^ arr[i]] > dp[k])
	      dp[k ^ arr[i]] = dp[k] + 1;

	    res[k ^ arr[i]] = 1;
	  }
      }

    int j = M - 1, k = n >> 1;
    while (dp[j] > k)
      --j;
    System.out.println (j);
  }
}

PHYTHON

from itertools import combinations


def fun(arr, N):
   sub = []
   max_xor = max(arr)
   for i in range(1, N // 2):
       comb = combinations(arr, i + 1)
       for i in comb:
           sub.append(list(i))
   for a in sub:
       z = 0
       for b in a:
           z = z ^ b
       if z > max_xor:
           max_xor = z
   return max_xor


N = int(input("Enter Length : "))
arr = []
for i in range(N):
   arr.append(int(input(f"Enter {i+1} Element : ")))
if N < 3:
   print("Output :", max(arr))
else:
   print("Output :", fun(arr, N))

C++

#include<bits/stdc++.h>
using namespace std;

int main ()
{
  int n;
  cin >> n;
  
  int arr[n];
  
  for (int i = 0; i < n; i++) cin >> arr[i];
  
  int M = 1 << 20;
  int dp[M];
  char res[M];
  
  for (int i = 1; i < M; i++)
    dp[i] = INT_MAX;
  
  for (int i = 0; i < n; i++)
    {
      if (arr[i] == 0)
	   continue;
      
      for (int j = 0; j < M; j++)
	    res[j] = 0;
      
      for (int k = 0; k < M; k++) { if (res[k] == 1) continue; if (dp[k] > dp[k ^ arr[i]])
	       dp[k] = dp[k ^ arr[i]] + 1;
	  
         else if (dp[k ^ arr[i]] > dp[k])
	       dp[k ^ arr[i]] = dp[k] + 1;
	     
	     res[k ^ arr[i]] = 1;
	      
	  }
    }
  
 int j = M - 1, k = n >> 1;
  
  while (dp[j] > k)
    j--;
  
  cout << j;
  
  return 0;

}

JAVA 

import java.util.*;
class Main
{
  public static boolean convertBase (int m, int base)
  {

    int rem = m % base;
      m = m / base;

    while (m >= base && (m % base == rem))
      m = m / base;

    if (m == rem)
      return true;

      return false;
  }

  public static void main (String[]args)
  {
    Scanner sc = new Scanner (System.in);
    int m = sc.nextInt ();
    int base = 2;

    while (convertBase (m, base) != true)
      base++;

    System.out.println (base);

  }

}

PHYTHON

def convertBase(m, base):
   rem = m % base
   m = m // base
   while m >= base and (m % base == rem):
       m = m // base
   if m == rem:
       return True
   return False


m = int(input())
base = 2
while not convertBase(m, base):
   base = base + 1
print(base)

C++

#include<bits/stdc++.h>
using namespace std;


bool converted (int M, int base)
{
  int rem = M % base;
  M /= base;

  while (M >= base)
    {
      if (M % base != rem)
	return 0;

      M /= base;
    }

  if (M == rem)
    return 1;

  return 0;
}

int main ()
{
  int M;
  cin >> M;

  int base = 2;

  while (converted (M, base) != 1)
    {
      base++;
    }

  cout << base;

  return 0;
}

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