IBM Associate System Engineer Coding Questions & Answers 2024 (C/C++/Java/Python) | IBM ASE MCQ Coding Questions 2024

IBM Associate System Engineer Coding Questions & Answers 2024 | IBM Associate System Engineer Coding Questions With Solution | IBM Associate System Engineer Coding Questions in Java | IBM Associate System Engineer Coding Questions in Python

IBM Associate System Engineer Coding Questions & Answers 2024: All of the fundamental algorithms, data structures, and programming concepts are included in the IBM Associate System Engineer Coding Questions. If you’re interested in working at IBM Associate System Engineer and don’t know how to prepare for the questions in the IBM Associate System Engineer Coding Round, we can help. The coding round is one of the most crucial parts of the IBM Associate System Engineer test. In this article, we covered a variety of sample IBM Associate System Engineer Coding Questions and potential solutions. We will talk about IBM Associate System Engineer Coding Questions in the below section

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IBM Associate System Engineer Coding Questions and Solutions 2024

Question) Henry is extremely keen on history and every one of the ages of his family. He does a ton of exploration and understands that he has plummeted from the incomparable Amaya line. After a ton of looking through old records and the most recent records of the general public, he can discover all the parent-kid connections in his family right from the extraordinary ruler Ming of the tradition to himself.

These connections are given in the structure of a direct exhibit where the emperor is at the main position and his kids are at pos (2i + 1) and (2i + 2)

This is the pattern followed throughout.

Henry needs to sort out every one of the kin of individual X from the information.

Write a program for Henry to figure out all the siblings of person X from the data.
Return the sorted list of all of Henry’s siblings.

If no sibling return {-1}

• input 1: N, the length of the array
• input2: An array representing the ancestral tree
• input 3: X, the person whose siblings are sought.
• output – return the array of all siblings in increasingly sorted order.

Example 1 :

input 1: 5
input 2: {1,2,3,4,5}
input 3: 1
output: {-1}
Explanation: x is the root of the tree and has no siblings

Example 2 :

input 1: 6
input 2: {1,2,3,4,5,6}
input 3: 5
output: {4,6}
Explanation: {2,3 } are the children of {1}. {4,5,6 } are the children of {2,3}, thus the siblings of x= 5 are {4,6}

JAVA PROGRAM

import java.util.*;
 
public class Main
{
    Node root;
    static class Node
    {
        int data;
        Node left, right;
    
        Node (int data) 
        {
            this.data = data;
            this.left = null;
            this.right = null;
        } 
    } 

    public static void main (String[]args)
    {
        Main t2 = new Main ();
        Scanner sc = new Scanner (System.in);
        int length = sc.nextInt ();
        int arr[] = new int[length];
    
        for (int i = 0; i < length; i++)
        {
            arr[i] = sc.nextInt ();
        }
        int target = sc.nextInt ();
        t2.root = t2.insertLevelOrder (arr, t2.root, 0);
    
        Set < Integer > sets = t2.levelOrderBottom (t2.root, target);
        sets.remove (target);
        System.out.println (sets);
    }   
    public static Node insertLevelOrder (int[]arr, Node root, int i) 
    {
        if (i < arr.length)
        {
            Node temp = new Node (arr[i]);
            root = temp;
            root.left = insertLevelOrder (arr, root.left, 2 * i + 1);
            root.right = insertLevelOrder (arr, root.right, 2 * i + 2);
        }
        return root;
    }

    public Set < Integer > levelOrderBottom (Node root, int target)
    {
        if (root == null)
        {
        	return null;
        }
        Queue < Node > q = new LinkedList <> ();
        q.offer (root);
    
        while (!q.isEmpty ())
        {
            int qsize = q.size ();
            Set < Integer > temp = new HashSet <> ();
            
            for (int i = 0; i < qsize; i++)
	        {
                Node child = q.poll ();
                temp.add (child.data);
                
                if (child.left != null)
	            {
                    q.offer (child.left);
                }
                if (child.right != null)
	            {
                    q.offer (child.right);
                }
            }
            if (temp.contains (target))
                return temp;
        }
        return null;
    }
}

PYTHON PROGRAM

def identify_siblings(tree_array, x):
    tree_len = len(tree_array)
    index = tree_array.index(x)
    level = 0
    start_index = level
    number_of_nodes = 0
    while start_index < tree_len:
    	end_index = pow(2, level) + start_index
    	if x in tree_array[start_index:end_index]:
    		break
    	level += 1
    	start_index = (2 * start_index) + 1
    final_array = tree_array[start_index:end_index]
    final_array.remove(x)
    return final_array if final_array else [-1]

n = int(input())
arr = []
for i in range(n):
    arr.append(int(input()))
x = int(input())
print(identify_siblings(arr, x))

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int N, ans, price;
    cin>>N;
   
    int arr[N];
    
    for (int i = 0; i < N; i++)
        cin>>arr[i];
    
    price = 0;
    ans = 0;
    for (int i = 0; i < N; i++)
    {
        price = price + arr[i];
        if (ans > price)
            {
                ans = price;
            }
    }
    cout << ans;
}

C PROGRAM:

#include<stdio.h>
int main()
{
    int N, ans, price;
    scanf("%d", &N);
   
    int arr[N];
    
    for (int i = 0; i < N; i++)
        scanf("%d", &arr[i]);
    
    price = 0;
    ans = 0;
    for (int i = 0; i < N; i++)
    {
        price = price + arr[i];
        if (ans > price)
            {
                ans = price;
            }
    }
    printf("%d", ans);
}

PYTHON PROGRAM

n = int(input())
arr = []
for i in range(n):
   arr.append(int(input()))
min = 0
for i in range(1, n):
   if sum(arr[:i]) < min:
       min = sum(arr[:i])
print(min)

Question) Given a positive whole number n, find the smallest number which has the very same digits existing in the whole number n and is greater than n. In the event that no such certain number exists, return – 1.

Note: that the returned number should fit in a 32-digit number, if there is a substantial answer however it doesn’t fit in a 32-bit number, return – 1.

Example 1:

Input: n = 12
Output: 21
Explanation: Using the same digit as the number of permutations, the next greatest number for 12 is 21.

Example 2:

Input: n = 21
Output: -1
Explanation: The returned integer does not fit in a 32-bit integer

C++ PROGRAM

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

void swap(char *a, char *b)
{
	char temp = *a;
	*a = *b;
	*b = temp;
}

void findNext(char number[], int n)
{
	int i, j;

	for (i = n-1; i > 0; i--)
		if (number[i] > number[i-1])
		break;

	if (i == 0)
	{
		cout << "Next number is not possible";
		return;
	}

	int x = number[i-1], smallest = i;
	for (j = i+1; j < n; j++) if (number[j] > x && number[j] < number[smallest])
			smallest = j;

	swap(&number[smallest], &number[i-1]);


	sort(number + i, number + n);

	cout << number; return; } int main() { char digits[]; cin >> digits
	int n = strlen(digits);
	findNext(digits, n);
	return 0;
}

JAVA PROGRAM

import java.util.*;
public class Main
{
	static void swap(char ar[], int i, int j)
	{
		char temp = ar[i];
		ar[i] = ar[j];
		ar[j] = temp;
	}
	
	static void findNext(char ar[], int n)
	{
		int i;
		
		for (i = n - 1; i > 0; i--)
		{
			if (ar[i] > ar[i - 1]) {
				break;
			}
		}
		
		if (i == 0)
		{
			System.out.println("Not possible");
		}
		else
		{
			int x = ar[i - 1], min = i;
			
			
			for (int j = i + 1; j < n; j++)
			{
				if (ar[j] > x && ar[j] < ar[min])
				{
					min = j;
				}
			}
		
			swap(ar, i - 1, min);
			Arrays.sort(ar, i, n);
			
			for (i = 0; i < n; i++)
				System.out.print(ar[i]);
		}
	}

	public static void main(String[] args)
	{
	    Scanner sc = new Scanner(System.in);
	    int n =sc.nextInt();
	    String s = sc.next();
		char digits[] = s.toCharArray();
		
		findNext(digits, n);
	}
}

PYTHON PROGRAM

arr = []
def permute(s,ans):
    n = len(s)
    if (n == 0):
        arr.append(ans)
        return
    for i in range(n):
        ch = s[i]
        L = s[0:i]
        R = s[i + 1:]
        REM = L+ R
        permute(REM,ans + ch)
ans= ""
s = input()
n = len(s)
permute(s,ans)
arr = list(set(arr))
arr.sort()
ind = arr.index(s)
if(ind == len(arr)-1):
    print(-1)
else:
    print(arr[ind+1])

IBM Associate System Engineer Coding Questions and Solutions in JAVA

Question) Rohan and his team are participating in the Treasure Hunt event of college in which in each step they have to solve one problem to get a clue about the Treasure location. Rohan’s team has performed very well and reached the final step where they have to provide a code of a problem to get a final clue about treasure .

Given a string s, they need to find the longest palindromic subsequence’s length in s.
A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

The string contains only lowercase letters.

Write a program to help Rohan’s team that takes in input as String x and returns the length of the longest palindromic subsequence of x.

Input Specification:

input1: string input

Output Specification:

Return the length of the longest palindromic subsequence

Example 1:

Input: s = “bbbab”
Output: 4
Explanation: One possible longest palindromic subsequence is “bbbb”.

Example 2:

Input: s = “cbbd”
Output: 2
Explanation: One possible longest palindromic subsequence is “bb”.

C++ PROGRAM

#include<iostream>
#include<cstring>
using namespace std;

int max (int x, int y) 
{
    return (x > y)? x : y; 
}

int lps(char *str)
{
    int n = strlen(str);
    int i, j, cl;
    int L[n][n]; 

    for (i = 0; i < n; i++)
	    L[i][i] = 1;

    for (cl = 2; cl <= n; cl++)
	{
		for (i = 0; i < n-cl+1; i++)
		{
		    j = i + cl-1;
		    if (str[i] == str[j] && cl == 2)
		        L[i][j] = 2;
		    else if (str[i] == str[j])
		        L[i][j] = L[i+1][j-1] + 2;
		    else L[i][j] = max(L[i][j-1], L[i+1][j]);
		}
	}
	return L[0][n-1];
}

int main()
{
    char seq[20];
    cin >> seq;
	int n = strlen(seq);
	printf("%d", lps(seq));
	getchar();
	return 0;
}

JAVA PROGRAM

import java.util.*;
public class Main
{
	static int max (int x, int y) 
	{
	    return (x > y)? x : y; 
	    
	}

	static int lps(String seq)
	{
	    int n = seq.length();
	    int i, j, cl;
    	    int L[][] = new int[n][n];
	
    	    for (i = 0; i < n; i++)
	    	L[i][i] = 1;
		
		for (cl = 2; cl <= n; cl++)
		{
			for (i = 0; i < n-cl+1; i++)
			{
				j = i+cl-1;
				if (seq.charAt(i) == seq.charAt(j) && cl == 2)
    				L[i][j] = 2;
				else if (seq.charAt(i) == seq.charAt(j))
	    			L[i][j] = L[i+1][j-1] + 2;
				else
		    		L[i][j] = max(L[i][j-1], L[i+1][j]);
			}
		}
		return L[0][n-1];
	}
	
	public static void main(String args[])
	{
	        Scanner sc = new Scanner(System.in);
		String seq = sc.next();
		int n = seq.length();
		System.out.println(""+ lps(seq));
	}
}

PYTHON PROGRAM

s = input()
n = len(s)
rev = s[::-1]
dp = [[0 for _ in range(n+1)] for _ in range(n+1)]

for i in range(1, n+1):
    for j in range(1, n+1):
        if s[i-1] == rev[j-1]:
            dp[i][j] = 1 + dp[i-1][j-1]
        else:
            dp[i][j] = max(dp[i-1][j], dp[i][j-1])

print(dp[n][n])

Question) Given a number n, the task is to find the remainder when n is divided by 11. The input number may be very large.

Since the given number can be very large, you can not use n % 11.

Input Specification:

inputs a large number in the form of a string

Output Specification:

Return the remainder modulo 11 of input1

Example1:

Input: str = 13589234356546756
Output: 6

Example2:

Input: str = 3435346456547566345436457867978
Output: 4

Example3:

input: str = 121
Output: 0

JAVA PROGRAM

import java.util.*;
import java.io.*;
public class Main
{
	static int remainder(String str)
	{
		int len = str.length();
		int num, rem = 0;
		
		for (int i = 0; i < len; i++)
		{
			num = rem * 10 + (str.charAt(i) - '0');
			rem = num % 11;
		}
		return rem;
	}
	
	public static void main(String args[])
	{
	    Scanner sc = new Scanner(System.in);
		String str = sc.next();
		System.out.println(remainder(str));
	}
}

PYTHON PROGRAM

s, rem = input(), 0

for i in s:
    rem = (rem * 10 + int(i)) % 11

print(rem)

IBM Associate System Engineer Coding Questions and Solutions in Python

Question) Williamson is an analyst he needs to analyse a particular topic for performing analysis for that he needs to find a permutation of a given object.He don’t know how to find permutation so for simplifying his work he is hiring one software developer who can code for him and find a permutation and combination of a given object.

Consider you are giving an interview to williamson for working with him. Find a permutation of given input to proof that you are fit for his requirement.

Input Specification:
nCr where n and r are numbers given by Williamson to you
nCr is defined as n! / (r! x (n-r)!)

Here, n! denotes the factorial of a number. Also, you have to calculate this number as modulo
Input Specification:

• input1: The number n.
• Input2: The number r.
• Input3: The number m,
  .

Output specification:

• The value of nCr%m.

Example 1:
Input1: 3
Input2: 2
Input3: 10000000009

Output:3

Explanation:
n=3.r=2,m=100 So, n!=3!=6, r!=2!=2, (n-1)!= 1!=1.
So, nCr = (6/(2*1))%1000000009= 3.

Example 2:
input1: 5
input2: 2
input3: 1000000009

Output: 10

Explanation:
n=5,r=2, m=100 So, n!=5!=120, r!=2=2, (n-1)!= 3!=6.
So, nCr = (120/(2*6))%1000000009= 10.

JAVA PROGRAM

import java.util.*;
public class Main
{
    static int fact (int number)
    {
        int f = 1;
        int j = 1;

        while (j <= number)
        {
            f = f * j;
            j++;
        }
        return f;
    }
    
    public static void main (String args[])
    {
        Scanner sc = new Scanner (System.in);
        int n = sc.nextInt ();
        int r = sc.nextInt ();
        long m = sc.nextLong ();
        int result = fact (n) / (fact (r) * fact (n - r));
        long finalresult = (long) result % m;
        System.out.println ("" + result);
    } 
}

PYTHON PROGRAM

import math
n=int(input())
r=int(input())
m=int(input())
#nCr= (n!)/((n-r)!)*(r!)
numerator=(math.factorial(n))
denominator=(math.factorial(n-r))*math.factorial(r)
ncr=numerator//denominator
print(ncr%m)

Question) A Derangement is a permutation of n elements, such that no element appears in its original position.
For example, a derangement of {0, 1, 2, 3} is {2, 3, 1, 0}.
Given a number n, find the total number of Derangements of a set of n elements.

Input Specification:
input1: N, the number of Objects

Output Specification:
Return the number of arrangements in which no object occurs at its original
position

Example 1:

Input: n = 2
Output: 1
For two elements say {0, 1}, there is only one possible derangement {1, 0}

Example 2:

Input: n = 3
Output: 2
For three elements say {0, 1, 2}, there are two possible derangements {2, 0, 1} and {1, 2, 0}

Example 3:

Input: n = 4
Output: 9
For four elements say {0, 1, 2, 3}, there are 9 possible derangements {1, 0, 3, 2} {1, 2, 3, 0} {1, 3, 0, 2}, {2, 3, 0, 1}, {2, 0, 3, 1}, {2, 3,1, 0}, {3, 0, 1, 2}, {3, 2, 0, 1} and {3, 2, 1, 0

C++ PROGRAM

#include <iostream>
using namespace std;

int countDer(int n)
{

	// base case
	if (n == 1 or n == 2) {
		return n - 1;
	}

	int a = 0;
	int b = 1;

	// using above recursive formula
	for (int i = 3; i <= n; ++i) {
		int cur = (i - 1) * (a + b);
		a = b;
		b = cur;
	}

		return b;
}

// Driver Code
int main()
{
	int n;
	cin>>n;
	cout  << countDer(n);
	return 0;
}

JAVA PROGRAM

import java.util.*;
import java.io.*;
public class Main {

    static int countDer(int n) {
        
        if(n == 1 || n == 2) {
            return n-1;
        }
    
    
        int a = 0;
        int b = 1;
    
        
        for (int i = 3; i <= n; ++i) {
            int cur = (i-1)*(a+b);
            a = b;
            b = cur;
        }
            
    
        return b;
    }
    
    public static void main (String[] args)
    {
        Scanner sc  = new Scanner(System.in);
        int n = sc.nextInt();
        System.out.println(""+countDer(n));
    
    }
}

IBM Associate System Engineer Coding Questions and Solutions in C++

Question) After Watching a movie at PVR, Adil is pondering over the number of ways in which he can pay for the movie. He has x1, x2, x3, x4 coins of values 1,2,5 and 10 respectively. He wants to determine the number of ways in which he can pay an amount A.

You need to fill in a function that returns the number of ways to pay total amount

Input Specifications:

Input 1: An integer value denoting the total amount to be paid

Output Specification:
Return an Integer value denoting the number of ways to pay the total amount

Example1:
Input1: 40
Output : 195

Example2:
Input1: 4
Output : 3

C++ PROGRAM

#include <iostream>
#include <climits>
using namespace std;

int coinsExchange(int amt, int *deno, int n, int *dp) {
	// base case
	if (amt == 0) {
		return dp[amt] = 0;
	}
	if (dp[amt] != -1) {
		return dp[amt];
	}

	// recursive case
	int ans = INT_MAX;
	for (int i = 0 ; i < n ; i++) {
		if (amt >= deno[i]) {
			int sa = coinsExchange(amt - deno[i], deno, n, dp);
			if (sa != INT_MAX) {
				ans = min(sa + 1, ans);
			}
		}
	}

	return dp[amt] = ans;
}

int main() {

	int deno[4] = {1, 2, 5, 10};
	int amt;
	cin>>amt;
	int *dp = new int[amt + 1];
	for (int i = 0 ; i <= amt ; i++) {
		dp[i] = -1;
	}
	cout << coinsExchange(amt, deno, 4, dp) << endl;

	return 0;
}

JAVA PROGRAM

import java.io.*;
import java.util.*;
public class Coin {
    
    static int count( int S[], int m, int n )
    {
        
        if (n == 0)
            return 1;
        
        if (n < 0)
            return 0;
      
        if (m <=0 && n >= 1)
            return 0;
    
        return count( S, m - 1, n ) +
               count( S, m, n-S[m-1] );
    }
    
    public static void main(String[] args)
    {   Scanner sc = new Scanner(System.in);
        int total  = sc.nextInt();
        int arr[] = {1, 2, 5,10};
        int m = arr.length;
        System.out.println( count(arr, m, total));
    }
}

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