INFOSYS EXAM ANSWER

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Exam pattern:-

Infosys Syllabus 2021

Infosys has just changed their Infosys Syllabus 2021 for On and Off-campus Recruitment. For any company, we analyze each on-campus or off-campus drive to provide the most exact Infosys Online Test Pattern. As observed in Infosys Written Test Syllabus 2021, the questions are very tough for Infosys Test Pattern 2021. Below you will find the detailed Infosys Exam Pattern for the 2021 year.

Infosys Aptitude Test Syllabus

Logical Reasoning is the most difficult section asked in the Written Test Pattern for Infosys.

There are 5 sections in Infosys Online Test Paper

INFOSYS EXAM PATTERN :- Click here

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EXAM Answer's (9/1/22) morning slot

Coding Answer's

XOR elements

C++ language

Infosys

perfect_sum(arr, s, result) :

x = [0]*len(arr)

j = len(arr) - 1

while (s > 0) :

x[j] = s % 2

s = s // 2

j -= 1

sum = 0

for i in range(len(arr)) :

if (x[i] == 1) :

sum += arr[i]

if (sum == result) :

print("{",end="");

for i in range(len(arr)) :

if (x[i] == 1) :

print(arr[i],end= ", ");

print("}, ",end="")

def print_subset(arr, K) :

x = pow(2, len(arr))

for i in range(1, x):

perfect_sum(arr, i, K)

# Driver code

arr = [ ]

n = int(input("Enter length of array : "))

s=int(input("Enter sum : "))

for i in range(n):

ele=int(input("Enter element : "))

arr.append(ele)

print_subset(arr, s)

This one is for permutations question

Python

class nodes

{

// Function to return the value of

// Binomial Coefficient C(n, k)

static int binomialCoeff(int n, int k)

{

if (k > n)

return 0;

int res = 1;

// Since C(n, k) = C(n, n-k)

if (k > n - k)

k = n - k;

// Calculate the value of

// [n * (n - 1) *---* (n - k + 1)] /

// [k * (k - 1) * ... * 1]

for (int i = 0; i < k; ++i)

{

res *= (n - i);

res /= (i + 1);

}

return res;

}

// Driver Code

public static void main(String[] args)

{

int N = 5, M = 1;

int P = (N * (N - 1)) / 2;

System.out.println(binomialCoeff(P, M));

}

N No of Nodes Code ☑️

Python

max value stored Code

John sum array

perfect_sum(arr, s, result) :

x = [0]*len(arr)

j = len(arr) - 1

while (s > 0) :

x[j] = s % 2

s = s // 2

j -= 1

sum = 0

for i in range(len(arr)) :

if (x[i] == 1) :

sum += arr[i]

if (sum == result) :

print("{",end="");

for i in range(len(arr)) :

if (x[i] == 1) :

print(arr[i],end= ", ");

print("}, ",end="")

def print_subset(arr, K) :

x = pow(2, len(arr))

for i in range(1, x):

perfect_sum(arr, i, K)

# Driver code

arr = [ ]

n = int(input("Enter length of array : "))

s=int(input("Enter sum : "))

for i in range(n):

ele=int(input("Enter element : "))

arr.append(ele)

print_subset(arr, s)

class Main {

public static void main(String[] args) {

int[] numbers = {2, -9, 0, 5, 12, -25, 22, 9, 8, 12};

int sum = 0;

Double average;

// access all elements using for each loop

// add each element in sum

for (int number: numbers) {

sum += number;

}

// get the total number of elements

int arrayLength = numbers.length;

// calculate the average

// convert the average from int to double

average = ((double)sum / (double)arrayLength);

System.out.println("Sum = " + sum);

System.out.println("Average = " + average);

}

Ali army code

Java

class Main {

public static void main(String[] args) {

int[] numbers = {2, -9, 0, 5, 12, -25, 22, 9, 8, 12};

int sum = 0;

Double average;

// access all elements using for each loop

// add each element in sum

for (int number: numbers) {

sum += number;

}

// get the total number of elements

int arrayLength = numbers.length;

// calculate the average

// convert the average from int to double

average = ((double)sum / (double)arrayLength);

System.out.println("Sum = " + sum);

System.out.println("Average = " + average);

}

John has an array

n=int(input())

x=int(input())

y=int(input())

z=int(input())

a=[]

for i in range(n):

a.append(int(input()))

s=(x*1)+(y*2)+(z*3)

c=0

for i in a:

if i<=s:

c+=1

print(c)

Chocolate factory code

Python

Shopping list code

Python

10)

General ali code

11)

Longest length Code

12)

Chocolate Factory Code

private static int minTime(int[] rate, int[] available) {

int n = rate.length;

int time = 0;

while (n > 0) {

int ind = findMax(rate);

if(time!=0){

available[ind] += (time * rate[ind]);

}

time += available[ind];

rate[ind] = Integer.MIN_VALUE;

n--;

}

return (arraySum(available) % 1000000007);

}

private static int findMax(int[] rate) {

int max = Integer.MIN_VALUE;

int maxIndex = -1;

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

if (rate[i] > max) {

max = rate[i];

maxIndex = i;

}

return maxIndex;

}

private static int arraySum(int[] available){

int sum = 0;

for(int i : available){

sum+=i;

}

return sum;

}

Ali cycle code

14)

array AR of size N

perfect_sum(arr, s, result) :

x = [0]*len(arr)

j = len(arr) - 1

while (s > 0) :

x[j] = s % 2

s = s // 2

j -= 1

sum = 0

for i in range(len(arr)) :

if (x[i] == 1) :

sum += arr[i]

if (sum == result) :

print("{",end="");

for i in range(len(arr)) :

if (x[i] == 1) :

print(arr[i],end= ", ");

print("}, ",end="")

def print_subset(arr, K) :

x = pow(2, len(arr))

for i in range(1, x):

perfect_sum(arr, i, K)

# Driver code

arr = [ ]

n = int(input("Enter length of array : "))

s=int(input("Enter sum : "))

for i in range(n):

ele=int(input("Enter element : "))

arr.append(ele)

print_subset(arr, s)

15)

Python

def check(a,n)

if n==1:

return 1

a.sort()

c=0

i=0

while(i<n-1):

if a[i+1]-a[i]==1:

i=i+2

c=c+1

else:

i=i+1

if c:

return c

else:

return 1

n=int(input())

a=[]

for i in range(n):

a.append(int(input()))

print(check(a,n)

16)

Python

Ali army

max_li=[]

def longest_subarray(arr, d):

(i, j, e) = (0, 1, 0)

for i in range(d - 1):

if arr[i] == arr[i + 1]:

j += 1

else:

j = 1

if e < j:

e = j

return e

temp=li.copy()

for each in li:

max_li.append(longest_subarray(temp, len(temp)))

temp=li.copy()

while(True):

try:

temp.remove(each)

except:

break

print(max(max_li))

17)

Python

John Code

n=int(input())

x=int(input())

y=int(input())

z=int(input())

a=[]

for i in range(n):

a.append(int(input()))

s=(x*1)+(y*2)+(z*3)

c=0

for i in a:

if i<=s:

c+=1

print(c)

18)

Whales code

Python

def check(a,n)

if n==1:

return 1

a.sort()

c=0

i=0

while(i<n-1):

if a[i+1]-a[i]==1:

i=i+2

c=c+1

else:

i=i+1

if c:

return c

else:

return 1

n=int(input())

a=[]

for i in range(n):

a.append(int(input()))

print(check(a,n))

Infosys Exam Answer's afternoon slot (9/1/22)

def check(a,n)

if n==1:

return 1

a.sort()

c=0

i=0

while(i<n-1):

if a[i+1]-a[i]==1:

i=i+2

c=c+1

else:

i=i+1

if c:

return c

else:

return 1

n=int(input())

a=[]

for i in range(n):

a.append(int(input()))

print(check(a,n))

array one

Python

Max of binary South

Python

n=int(input())

m=int(input())

s=input()

t=input()

def subString(s,n):

c=0

for i in range(n):

for len in range(i+1,n+1):

a=s[i: len]

c+=1

return c-2

if s==t:

print(1)

else:

if t in s:

print(subString(s,n))

else:

print(0)

Python

Maximum possibility of unique digits in k divided subarrray of given arraylist

Java code

INFOSYS EXAM ANSWER'S (10/1/22)

Angry professor

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int t,count=0;
scanf("%d",&t);
for(int a0 = 0; a0 < t; a0++){
count=0;
int n;
int k;
scanf("%d %d",&n,&k);
int a[n];
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}

for(int a_i = 0; a_i < n; a_i++){
if(a[a_i]<=0)
count++;
}
if(count<k)
printf("YES\n");
else
printf("NO\n");
//count=0;
}
return 0;
}

C language & python

Permutations code

Minimum swaps

C language

South Indian kingdom

Java

import java.util.*;

class abc {

public static void main(String[] args) {

List<String> chola = new ArrayList<String>();

List<String> chera = new ArrayList<String>();

List<String> pallava = new ArrayList<String>();

List<String> pandya = new ArrayList<String>();

List<String> renounce = new ArrayList<String>();

Scanner kb = new Scanner(System.in);

int q = kb.nextInt();

for (int i=0; i <q; i++)

{

String name = kb.next();

int x = kb.nextInt();

int y = kb.nextInt();

if (x == 0 || y == 0)

renounce.add(name);

else if(x >0 && y >0)

chola.add(name);

else if(x >0 && y <0)

chera.add(name);

else if(x <0 && y <0)

pallava.add(name);

else if(x <0 && y >0)

pandya.add(name);

}

System.out.println("chola");

System.out.println(chola);

System.out.println("chera");

System.out.println(chera);

System.out.println("pallava");

System.out.println(pallava);

System.out.println("pandya");

System.out.println(pandya);

System.out.println("renounce");

System.out.println(renounce);

}

Maximize number of groups formed with size not smaller than its largest element

C++

Output :- 2

// C++ program to find minimum time

// required to produce m items.

#include<bits/stdc++.h>

using namespace std;

// Return the minimum time required to

// produce m items with given machines.

int minTime(int arr[], int n, int m)

{

// Initialize time, items equal to 0.

int t = 0;

while (1)

{

int items = 0;

// Calculating items at each second

for (int i = 0; i < n; i++)

items += (t / arr[i]);

// If items equal to m return time.

if (items >= m)

return t;

t++; // Increment time

}

// allcoding1

int main()

{

int arr[] = { 1, 2, 3 };

int n = sizeof(arr)/sizeof(arr[0]);

int m = 11;

cout << minTime(arr, n, m) << endl;

return 0;

}

Minimum time

C language

Red And White Flower

Python Program

28/1/22 EXAM Answer's

Whales code

Python

def check(a,n)

if n==1:

return 1

a.sort()

c=0

i=0

while(i<n-1):

if a[i+1]-a[i]==1:

i=i+2

c=c+1

else:

i=i+1

if c:

return c

else:

return 1

n=int(input())

a=[]

for i in range(n):

a.append(int(input()))

print(check(a,n))

C++

* This C++ program tells if there exists a pair in array whose sum results in x.

#include <iostream>

using namespace std;

// Function to find and print pair

bool chkPair(int A[], int size, int x) {

for (int i = 0; i < (size - 1); i++) {

for (int j = (i + 1); j < size; j++) {

if (A[i] + A[j] == x) {

cout << "Pair with a given sum " << x << " is (" << A[i] << ", " << A[j] << ")"

<< endl;

return 1;

}

return 0;

}

int main(void) {

int A[] = {0, -1, 2, -3, 1};

int x = -2;

int size = sizeof(A) / sizeof(A[0]);

if (chkPair(A, size, x)) {

cout << "Valid pair exists" << endl;

}

else {

cout << "No valid pair exists for " << x << endl;

}

return 0;

}

def minOps(A, B):

m = len(A)

n = len(B)

# This part checks whether conversion is possible or not

if n != m:

return -1

count = [0] * 256

for i in range(n): # count characters in A

count[ord(B[i])] += 1

for i in range(n): # subtract count for every char in B

count[ord(A[i])] -= 1

for i in range(256): # Check if all counts become 0

if count[i]:

return -1

# This part calculates the number of operations required

res = 0

i = n-1

j = n-1

while i >= 0:

# if there is a mismatch, then keep incrementing

# result 'res' until B[j] is not found in A[0..i]

while i>= 0 and A[i] != B[j]:

i -= 1

res += 1

# if A[i] and B[j] match

if i >= 0:

i -= 1

j -= 1

return res

# Driver program

A = "EACBD"

B = "EABCD"

print ("Minimum number of operations required is " + str(minOps(A,B)))

# This code is contributed

#Minimum number of operations required

C++ program to count distinct elements

// in a given array

#include <iostream>

using namespace std;

int countDistinct(int arr[], int n)

{

int res = 1;

// Pick all elements one by one

for (int i = 1; i < n; i++) {

int j = 0;

for (j = 0; j < i; j++)

if (arr[i] == arr[j])

break;

// If not printed earlier, then print it

if (i == j)

res++;

}

return res;

}

// Driver program to test above function

int main()

{

int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };

int n = sizeof(arr) / sizeof(arr[0]);

cout << countDistinct(arr, n);

return 0;

}

Distinct elements in array code

1. Jolly num code

// Program for Jolly Jumper Sequence

#include<bits/stdc++.h>

using namespace std;

// Function to check whether given sequence is

// Jolly Jumper or not

bool isJolly(int a[], int n)

{

// Boolean vector to diffSet set of differences.

// The vector is initialized as false.

vector<bool> diffSet(n, false);

// Traverse all array elements

for (int i=0; i < n-1 ; i++)

{

// Find absolute difference between current two

int d = abs(a[i]-a[i+1]);

// If difference is out of range or repeated,

// return false.

if (d == 0 d > n-1 diffSet[d] == true)

return false;

// Set presence of d in set.

diffSet[d] = true;

}

return true;

}

// Driver Code

int main()

{

int a[] = {11, 7, 4, 2, 1, 6};

int n = sizeof(a)/ sizeof(a[0]);

isJolly(a, n)? cout << "Yes" : cout << "No";

return 0;

}

Jolly num code

INFOSYS EXAM ANS (23/4/22)

Infosys exam Answer

Maximize number of groups formed with size not smaller than its largest element

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