Minimum operations to make sum at least M from given two Arrays

19 June 2025

0

Given arrays A[] and B[] of size N and integer M, the task is to find out the minimum operations required to collect a sum of at least M by performing the following operations any number of times. Either choosing the first element of A[] or the first element of B[] remove that element from the front and add that to the sum. Initially, the sum is zero.

Examples:

Input: A[] = {1, 9, 1, 4, 0, 1}, B[] = {3, 2, 1, 5, 9, 10}, M = 12
Output: 3
Explanation: Initially sum = 0

Removing front element of array A[] which is 1 and adding it to sum, now sum = 1.

Removing front element of array A[] which is 9 again and adding it to sum, now sum = 10.

Removing front element of array B[] which is 3 and adding it to sum, now sum = 13.

As 13 is greater than equal to M. Hence in three operations sum greater than equal to M = 12 can be achieved.

Input: A[] = {0, 1, 2, 3, 5}, B[] = {5, 0, 0, 0, 9}, M = 6
Output: 3

Naive approach: The basic way to solve the problem is as follows:

The basic way to solve this problem is to generate all possible combinations by using a recursive approach.

Time Complexity: O(2^N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following idea:

Dynamic programming can be used to solve this problem

dp[i][j][k] represents a minimum number of operations to make at least i from the first j elements of array A[] and first k elements of array B[], Here i represents the amount to be made, j represents the j^thelement of A[] and k represents the k^th element of B[].

It can be observed that the recursive function is called exponential times. That means that some states are called repeatedly. So the idea is to store the value of each state.

This can be done using by the store the value of a state and whenever the function is called, returning the stored value without computing again.

Follow the steps below to solve the problem:

Create a recursive function that takes three parameters i represents the amount to be made, j represents the jth element of A[] and k represents k’th element of B[].
Create a 3d array dp[M][N][N] initially filled with -1.
If the answer for a particular state is computed then save it in dp[i][j][k].
if the answer for a particular state is already computed then just return dp[i][j][k].
Base case if i becomes zero return 0.
Call the recursive function for choosing both the jth element of A[] and k’th element of B[].
return the dp value by dp[i][j][k] = ans.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
 
// dp table initialized with -1
int dp[501][101][101];
 
// Recursive Function to minimize the
// operations to collect at least sum of M
int solve(int i, int j, int k, int A[], int B[], int N)
{
 
    // Base case
    if (i <= 0) {
        return 0;
    }
 
    // If answer for current state is
    // already calculated then just
    // return dp[i][j][k]
    if (dp[i][j][k] != -1)
        return dp[i][j][k];
 
    // Answer initialized with zero
    int ans = 1e9;
 
    // Calling recursive function for
    // taking j'th element of array A[]
    if (j != N)
        ans = min(ans,
                  solve(i - A[j], j + 1, k, A, B, N) + 1);
 
    // Calling recursive function for
    // taking k'th element of array B[]
    if (k != N)
        ans = min(ans,
                  solve(i - B[k], j, k + 1, A, B, N) + 1);
 
    // Save and return dp value
    return dp[i][j][k] = ans;
}
 
// Function to minimize the operations
// to collect at least sum of M
int minOperations(int A[], int B[], int N, int M)
{
 
    // Filling dp table with - 1
    memset(dp, -1, sizeof(dp));
 
    // Minimum operations
    int ans = solve(M, 0, 0, A, B, N);
 
    return ans;
}
 
// Driver Code
int main()
{
 
    // Input 1
    int A[] = { 1, 9, 1, 4, 0, 1 },
        B[] = { 3, 2, 1, 5, 9, 10 };
    int N = sizeof(A) / sizeof(A[0]);
    int M = 12;
 
    // Function Call
    cout << minOperations(A, B, N, M) << endl;
 
    // Input 2
    int A1[] = { 0, 1, 2, 3, 5 }, B1[] = { 5, 0, 0, 0, 9 };
    int N1 = sizeof(A1) / sizeof(A1[0]);
    int M1 = 6;
 
    // Function Call
    cout << minOperations(A1, B1, N1, M1) << endl;
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;
import java.util.*;
 
class GFG {
 
  // dp table initialized with -1
  static int[][][] dp = new int[501][101][101];
 
  // Recursive Function to minimize the
  // operations to collect at least sum of M
  static int solve(int i, int j, int k, int[] A, int[] B,
                   int N)
  {
 
    // Base case
    if (i <= 0) {
      return 0;
    }
 
    // If answer for current state is
    // already calculated then just
    // return dp[i][j][k]
    if (dp[i][j][k] != -1) {
      return dp[i][j][k];
    }
 
    // Answer initialized with zero
    int ans = (int)1e9;
 
    // Calling recursive function for
    // taking j'th element of array A[]
    if (j != N) {
      ans = Math.min(
        ans,
        solve(i - A[j], j + 1, k, A, B, N) + 1);
    }
 
    // Calling recursive function for
    // taking k'th element of array B[]
    if (k != N) {
      ans = Math.min(
        ans,
        solve(i - B[k], j, k + 1, A, B, N) + 1);
    }
 
    // Save and return dp value
    return dp[i][j][k] = ans;
  }
 
  // Function to minimize the operations
  // to collect at least sum of M
  static int minOperations(int[] A, int[] B, int N, int M)
  {
 
    // Filling dp table with - 1
    for (int i = 0; i < dp.length; i++) {
      for (int j = 0; j < dp[i].length; j++) {
        Arrays.fill(dp[i][j], -1);
      }
    }
 
    // Minimum operations
    int ans = solve(M, 0, 0, A, B, N);
 
    return ans;
  }
 
  public static void main(String[] args)
  {
    // Input 1
    int[] A = { 1, 9, 1, 4, 0, 1 },
    B = { 3, 2, 1, 5, 9, 10 };
    int N = A.length;
    int M = 12;
 
    // Function Call
    System.out.println(minOperations(A, B, N, M));
 
    // Input 2
    int[] A1 = { 0, 1, 2, 3, 5 }, B1
      = { 5, 0, 0, 0, 9 };
    int N1 = A1.length;
    int M1 = 6;
 
    // Function Call
    System.out.println(minOperations(A1, B1, N1, M1));
  }
}
 
// This code is contributed by lokesh.

Python3

# Python code for the above approach
import sys
from typing import List
# dp table initialized with -1
dp = [[[-1 for _ in range(101)] for _ in range(101)] for _ in range(501)]
 
def solve(i: int, j: int, k: int, A: List[int], B: List[int], N: int) -> int:
    # Base case
    if i <= 0:
        return 0
       
    # If answer for current state is already calculated then just return dp[i][j][k]
    if dp[i][j][k] != -1:
        return dp[i][j][k]
       
    # Answer initialized with big value
    ans = sys.maxsize
     
    # Calling recursive function for taking j'th element of array A[]
    if j != N:
        ans = min(ans, solve(i - A[j], j + 1, k, A, B, N) + 1)
         
    # Calling recursive function for taking k'th element of array B[]
    if k != N:
        ans = min(ans, solve(i - B[k], j, k + 1, A, B, N) + 1)
         
    # Save and return dp value
    dp[i][j][k] = ans
    return ans
 
def minOperations(A: List[int], B: List[int], N: int, M: int) -> int:
    # Minimum operations
    return solve(M, 0, 0, A, B, N)
 
# Driver code
A = [1, 9, 1, 4, 0, 1]
B = [3, 2, 1, 5, 9, 10]
N = len(A)
M = 12
print(minOperations(A, B, N, M))
 
A1 = [0, 1, 2, 3, 5]
B1 = [5, 0, 0, 0, 9]
N1 = len(A1)
M1 = 6
print(minOperations(A1, B1, N1, M1))
 
# This code is contributed by lokeshpotta20.

C#

// C# code to implement the approach
 
using System;
using System.Linq;
 
public class GFG {
 
    // dp table initialized with -1
    static int[, , ] dp = new int[501, 101, 101];
 
    // Recursive Function to minimize the
    // operations to collect at least sum of M
    static int Solve(int i, int j, int k, int[] A, int[] B,
                     int N)
    {
 
        // Base case
        if (i <= 0) {
            return 0;
        }
 
        // If answer for current state is
        // already calculated then just
        // return dp[i][j][k]
        if (dp[i, j, k] != -1) {
            return dp[i, j, k];
        }
 
        // Answer initialized with zero
        int ans = (int)1e9;
 
        // Calling recursive function for
        // taking j'th element of array A[]
        if (j != N) {
            ans = Math.Min(
                ans,
                Solve(i - A[j], j + 1, k, A, B, N) + 1);
        }
 
        // Calling recursive function for
        // taking k'th element of array B[]
        if (k != N) {
            ans = Math.Min(
                ans,
                Solve(i - B[k], j, k + 1, A, B, N) + 1);
        }
 
        // Save and return dp value
        return dp[i, j, k] = ans;
    }
 
    // Function to minimize the operations
    // to collect at least sum of M
    static int MinOperations(int[] A, int[] B, int N, int M)
    {
        // Filling dp table with - 1
        for (int i = 0; i < dp.GetLength(0); i++) {
            for (int j = 0; j < dp.GetLength(1); j++) {
                for (int k = 0; k < dp.GetLength(2); k++) {
                    dp[i, j, k] = -1;
                }
            }
        }
 
        // Minimum operations
        int ans = Solve(M, 0, 0, A, B, N);
 
        return ans;
    }
 
    static public void Main()
    {
 
        // Code
        // Input 1
        int[] A = { 1, 9, 1, 4, 0, 1 },
              B = { 3, 2, 1, 5, 9, 10 };
        int N = A.Length;
        int M = 12;
 
        // Function Call
        Console.WriteLine(MinOperations(A, B, N, M));
 
        // Input 2
        int[] A1 = { 0, 1, 2, 3, 5 }, B1
                                      = { 5, 0, 0, 0, 9 };
        int N1 = A1.Length;
        int M1 = 6;
 
        // Function Call
        Console.WriteLine(MinOperations(A1, B1, N1, M1));
    }
}
 
// This code is contributed by lokeshmvs21.

Javascript

// dp table initialized with -1
let dp = new Array(501);
for(let i = 0; i < 501; i++) {
    dp[i] = new Array(101);
    for(let j = 0; j < 101; j++) {
        dp[i][j] = new Array(101).fill(-1);
    }
}
 
// Recursive Function to minimize the
// operations to collect at least sum of M
function solve(i, j, k, A, B, N) {
 
    // Base case
    if (i <= 0) {
        return 0;
    }
 
    // If answer for current state is
    // already calculated then just
    // return dp[i][j][k]
    if (dp[i][j][k] !== -1)
        return dp[i][j][k];
 
    // Answer initialized with zero
    let ans = 1e9;
 
    // Calling recursive function for
    // taking j'th element of array A[]
    if (j !== N)
        ans = Math.min(ans,
                      solve(i - A[j], j + 1, k, A, B, N) + 1);
 
    // Calling recursive function for
    // taking k'th element of array B[]
    if (k !== N)
        ans = Math.min(ans,
                      solve(i - B[k], j, k + 1, A, B, N) + 1);
 
    // Save and return dp value
    return dp[i][j][k] = ans;
}
 
// Function to minimize the operations
// to collect at least sum of M
function minOperations(A, B, N, M) {
 
    // Minimum operations
    let ans = solve(M, 0, 0, A, B, N);
 
    return ans;
}
 
// Input 1
let A = [1, 9, 1, 4, 0, 1];
let B = [3, 2, 1, 5, 9, 10];
let N = A.length;
let M = 12;
 
// Function Call
console.log(minOperations(A, B, N, M));
 
// Input 2
let A1 = [0, 1, 2, 3, 5]; 
let B1 = [5, 0, 0, 0, 9];
let N1 = A1.length;
let M1 = 6;
 
// Function Call
console.log(minOperations(A1, B1, N1, M1));
 
// This code is contributed by aadityamaharshi21.

Output

3
3

Time Complexity: O(N² * M)
Auxiliary Space: O(N²* M)

Related Articles:

Introduction to Dynamic Programming – Data Structures and Algorithms Tutorials

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