**Huffman Coding Questions & Answers 2024 | Huffman Coding Questions With Solutions | Huffman Coding Questions | Huffman Coding Model Questions | Huffman Coding Solved Example | Huffman Coding Algorithm Questions**

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**Huffman Coding Questions **

- Execute this function: Int OperationsBinarystring(char* str)
- Execute this function: Def Autocount(x)
- Adam decides to do some charity work. From day 1 till day x, he will give i^2 coins to charity. On day ‘i’ (1 < = i < = x), find the number of coins he gives to charity.
- Find the sum of the divisors for the N integer number.
- Write a program that accepts the integer array of length ‘size’ and finds the largest number that can be formed by permutation.
- Form an array of 1000 integers and find out the second-largest number. If there is no second-largest number, return the value to –1.
- Execute this function: def differenceofSum(n.m)
- Execute this function: Int Numberofcarry(Integer num 1, Integer num 2)
- Execute this function: def LargeSmallSum(arr)
- Execute this function: def Productsmallpair(sum,arr)

**Huffman Coding Questions and Answers 2024**

**Questions:** 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.

**Output**

- 2

**Example Programs:**

**JAVA PROGRAM**

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); } }

**Questions: **One of the first lessons IT students learn is the representation of natural numbers in the binary number system (base 2) This system uses only two digits, 0 and 1. In everyday life we use for convenience the decimal system (base 10) which uses ten digits, from 0 to 9. In general, we could use any numbering system.

Computer scientists often use systems based on 8 or 16. The numbering system based on K uses K digits with a value from 0 to K-1. Suppose a natural number M is given, written in the decimal system To convert it to the corresponding writing in the system based on K, we successively divide M by K until we reach a quotient that is less than K

The representation of M in the system based on K is formed by the final quotient (as first digit) and is followed by the remainder of the previous divisions**For example :**

- If M=122 and K=8, 122 in base 10= 172 in base 8 This means that the number
- In decimal system = 172 in octal system.
- 172 in base 8 = 1*8^2 + 7*8 + 2 = 122

You made the following observation in applying the above rule of converting natural numbers to another numbering system

- In some cases in the new representation all the digits of the number are the same. For example 63 in base 10= 333 in base 4

Given a number M in its decimal representation, your task is find the minimum base B such that in the representation of M at base B all digits are the same.

**Input Format**

- The first line contains an integer, M, denoting the number given

**Constraints**

- 1 <= M = 10^12

**Sample Input 1 :**

41

**Sample Output 1 :**

40

**Explanation :**

Here 41 in base 40. will be 11 so it has all digits the same, and there is no smaller base satisfying the requirements

**Sample Input 2 :**

34430

**Sample Output 2 :**

312

**Explanation :**

Here 34430 in base 312 will have all digits the same and there is no smaller base satisfying the requirements

**Example Programs:**

**JAVA PROGRAM**

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); } }

**Question: **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.

**EXAMPLE PROGRAMS**

**C++ PROGRAM**

#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 PROGRAM**

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); } }

**Questions: **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.

**EXAMPLE PROGRAMS**

**C++ PROGRAM**

#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; }

**PYTHON PROGRAM**

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))

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