Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

Count the combination of 4s and/or 5s required to make each Array element 0

  • Difficulty Level : Medium
  • Last Updated : 29 Jul, 2021

Given an array arr[] consisting of N positive integers, the task is to check if each array element can be represented in terms of numbers 4 and 5. If it is possible, then print the total count of 4 and 5 needed for each array element. Otherwise, print -1.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = {12, 11, 9}
Output: 3 -1 2
Explanation: 



  1. arr[0]( =12): Print 3 as it can be represented by using 3 fours i.e. 4 + 4 + 4 = 12.
  2. arr[1](= 11): Print -1 as it cannot be represented by any combination of 4 and 5.
  3. arr[2](= 9): Print 2 as it can be represented by using one 4 and one 5 i.e. 4 + 5 = 9.

Input: arr[] = {7, 15, 17, 22}
Output: -1 3 4 5

Approach: The problem can be solved using greedy approach and a bit of mathematics. Follow the steps below to solve the problem:

  • Initialize a vector say ans, as {-1} where the ans[i] stores the answer to a possible answer for the value arr[i] of the array arr[].
  • Iterate over a range [0, N-1] using the variable i and perform the following steps:
    • If arr[i] is less than 4, then, continue.
    • Initialize two variables, say sum as INT_MAX to store the count of numbers of 4 and 5 needed to form arr[i] and cnt as 0 to keep the count of the current factor of 4.
    • Iterate over a range [0, arr[i]] using the variable j and perform the following steps:
      • If (arr[i] – j) is divisible by 5, then, set the value of the sum to the minimum of sum and (cnt + (arr[i] – j)/5).
      • Increase the value of cnt by 1 and j by 4.
    • If the sum is not equal to INT_MAX, then, set the value of ans[i] in the vector ans as sum.
  • Finally, after completing the above steps, print the values in the vector ans.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the count of the
// combination of 4 or 5 required to
// make the arr[i] for each 0 < i < N
void sumOfCombinationOf4OR5(vector<int> arr, int N)
{
 
    // Vector to store the answer
    vector<int> ans(N, -1);
 
    // Iterate in the range[0, N-1]
    for (int i = 0; i < N; i++) {
 
        if (arr[i] < 4) {
            continue;
        }
 
        // Initialize sum to store the count
        // of numbers and cnt for the current
        // factor of 4
        int sum = INT_MAX, cnt = 0;
 
        // Iterate in the range[0, arr[i]] with
        // increment of 4
        for (int j = 0; j <= arr[i]; j += 4) {
 
            // Check if arr[i] - j(the current factor
            // of 4) is divisible by 5 or not
            if ((arr[i] - j) % 5 == 0) {
                sum = min(sum, cnt + (arr[i] - j) / 5);
            }
 
            cnt++;
        }
 
        // If sum is not maximum
        // then answer is found
        if (sum != INT_MAX)
            ans[i] = sum;
    }
 
    // Finally, print the required answer
    for (auto num : ans)
        cout << num << " ";
}
 
// Driver Code
int main()
{
 
    // Given Input
    vector<int> arr = { 7, 15, 17, 22 };
    int N = arr.size();
 
    // Function Call
    sumOfCombinationOf4OR5(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.Arrays;
 
class GFG
{
   
    // Function to print the count of the
    // combination of 4 or 5 required to
    // make the arr[i] for each 0 < i < N
    static void sumOfCombinationOf4OR5(int[] arr, int N)
    {
 
        // Vector to store the answer
        int[] ans = new int[N];
        Arrays.fill(ans, -1);
        // Iterate in the range[0, N-1]
        for (int i = 0; i < N; i++) {
 
            if (arr[i] < 4) {
                continue;
            }
 
            // Initialize sum to store the count
            // of numbers and cnt for the current
            // factor of 4
            int sum = Integer.MAX_VALUE;
            int cnt = 0;
 
            // Iterate in the range[0, arr[i]] with
            // increment of 4
            for (int j = 0; j <= arr[i]; j += 4) {
 
                // Check if arr[i] - j(the current factor
                // of 4) is divisible by 5 or not
                if ((arr[i] - j) % 5 == 0) {
                    sum = Math.min(sum,
                                   cnt + (arr[i] - j) / 5);
                }
 
                cnt++;
            }
 
            // If sum is not maximum
            // then answer is found
            if (sum != Integer.MAX_VALUE)
                ans[i] = sum;
        }
 
        // Finally, print the required answer
        for (int num : ans)
            System.out.printf(num + " ");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Given Input
        int[] arr = { 7, 15, 17, 22 };
        int N = arr.length;
 
        // Function Call
        sumOfCombinationOf4OR5(arr, N);
    }
}
 
// This code is contributed by Potta Lokesh

Python3




# Python3 program for the above approach
 
# Function to print the count of the
# combination of 4 or 5 required to
# make the arr[i] for each 0 < i < N
def sumOfCombinationOf4OR5(arr, N):
 
    # Vector to store the answer
    ans = [-1 for i in range(N)]
 
    # Iterate in the range[0, N-1]
    for i in range(N):
         
        if (arr[i] < 4):
            continue
 
        # Initialize sum to store the count
        # of numbers and cnt for the current
        # factor of 4
        sum = 10**9
        cnt = 0
 
        # Iterate in the range[0, arr[i]] with
        # increment of 4
        for j in range(0, arr[i] + 1, 4):
 
            # Check if arr[i] - j(the current factor
            # of 4) is divisible by 5 or not
            if ((arr[i] - j) % 5 == 0):
                sum = min(sum, cnt + (arr[i] - j) // 5)
 
            cnt += 1
 
        # If sum is not maximum
        # then answer is found
        if (sum != 10**9):
            ans[i] = sum
 
    # Finally, print the required answer
    for num in ans:
        print(num, end = " ")
 
# Driver Code
 
# Given Input
arr = [ 7, 15, 17, 22 ]
N = len(arr)
 
# Function Call
sumOfCombinationOf4OR5(arr, N)
 
# This code is contributed by gfgking

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to print the count of the
// combination of 4 or 5 required to
// make the arr[i] for each 0 < i < N
static void sumOfCombinationOf4OR5(int []arr, int N)
{
 
    // Vector to store the answer
    int []ans = new int[N];
    for(int i = 0; i < N; i++)
       ans[i] = -1;
    
 
    // Iterate in the range[0, N-1]
    for (int i = 0; i < N; i++) {
 
        if (arr[i] < 4) {
            continue;
        }
 
        // Initialize sum to store the count
        // of numbers and cnt for the current
        // factor of 4
        int sum = Int32.MaxValue, cnt = 0;
 
        // Iterate in the range[0, arr[i]] with
        // increment of 4
        for (int j = 0; j <= arr[i]; j += 4) {
 
            // Check if arr[i] - j(the current factor
            // of 4) is divisible by 5 or not
            if ((arr[i] - j) % 5 == 0) {
                sum = Math.Min(sum, cnt + (int)(arr[i] - j) / 5);
            }
 
            cnt++;
        }
 
        // If sum is not maximum
        // then answer is found
        if (sum != Int32.MaxValue)
            ans[i] = sum;
    }
 
    // Finally, print the required answer
    foreach(int num in ans)
        Console.Write(num + " ");
}
 
// Driver Code
public static void Main()
{
 
    // Given Input
    int []arr = {7, 15, 17, 22 };
    int N = arr.Length;
 
    // Function Call
    sumOfCombinationOf4OR5(arr, N);
}
}
 
// This code is contributed by ipg2016107.

Javascript




<script>
// Javascript program for the above approach
 
// Function to print the count of the
// combination of 4 or 5 required to
// make the arr[i] for each 0 < i < N
function sumOfCombinationOf4OR5(arr, N) {
 
    // Vector to store the answer
    let ans = new Array(N).fill(-1);
 
    // Iterate in the range[0, N-1]
    for (let i = 0; i < N; i++) {
 
        if (arr[i] < 4) {
            continue;
        }
 
        // Initialize sum to store the count
        // of numbers and cnt for the current
        // factor of 4
        let sum = Number.MAX_SAFE_INTEGER, cnt = 0;
 
        // Iterate in the range[0, arr[i]] with
        // increment of 4
        for (let j = 0; j <= arr[i]; j += 4) {
 
            // Check if arr[i] - j(the current factor
            // of 4) is divisible by 5 or not
            if ((arr[i] - j) % 5 == 0) {
                sum = Math.min(sum, cnt + Math.floor((arr[i] - j) / 5));
            }
 
            cnt++;
        }
 
        // If sum is not maximum
        // then answer is found
        if (sum != Number.MAX_SAFE_INTEGER)
            ans[i] = sum;
    }
 
    // Finally, print the required answer
    for (let num of ans)
        document.write(num + " ");
}
 
// Driver Code
 
 
// Given Input
let arr = [7, 15, 17, 22];
let N = arr.length;
 
// Function Call
sumOfCombinationOf4OR5(arr, N);
</script>
 
// This code is contributed by _saurabh_jaiswal.
Output
-1 3 4 5 

Time Complexity: O(N*M) where M is the maximum element of the array arr[].
Auxiliary Space: O(N)

Note: The above approach can be further optimized in terms of space complexity as storing the result before printing is optional in the above approach. Thereafter, the space complexity will be optimized to O(1).




My Personal Notes arrow_drop_up