Minimize transfer operations to get the given Array sum
Last Updated :
23 Jul, 2025
Given two integer arrays A[] and B[] of length N and M respectively, the task is to find the minimum number of operations required such that the sum of elements in array A becomes targetSum. In one operation you can transfer an element from A[] to B[] or from B[] to A[].
Examples
Input: N = 4, M = 3, targetSum = 12, A[] = {1, 2, 1, 4}, B[] = {5, 12, 11}
Output: 2
Explanation: We can do the following two operations
1) Transfer 1 from A[] to B[]
2) Transfer 5 from B[] to A[]
So, the array becomes 5 + 2 + 1 + 4 = 12, which is equal to the target sum.
Input: N = 4, M = 5, targetSum = 12, A[] = {7, 2, 4, 3], B[] = {8, 1, 1, 7, 9}
Output: 1
Explanation: We can do the following two operations
1) Transfer 4 from A[] to B[]
So, the array becomes 7 + 2 + 3 = 12, which is equal to the target sum.
Approach: Implement the idea below to solve the problem:
We can reduce this problem into subproblems. Let's say elements with sum p are transferred from array A[] to array B[] and elements with sum q are transferred from array B[] to array A[]. So, the number of operations will be minimum when we take minimum number of elements from array A[] with sum p and minimum number of elements from array B[] with sum q.
This Problem can be solved by using Dynamic Programming, where we have to find minimum number of elements in an array with a given sum.
Follow the below steps to implement the idea:
- Create two dynamic arrays (say dp1[][] and dp2[][]).
- Let dp1[i][j] represent the minimum number of elements required in array A to make sum j till index i - 1, similarly, dp2[i][j] can also be defined for array B[] in the same way.
- We can create dp1[][] and dp2[][] using this transformation dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - a[i - 1]] + 1).
- The final value of array A[] will be targetSum, let p be the sum removed from array A[] and q be the sum removed from array B[](added to array A). Then the total number of operations will be dp1[N][p]+dp2[M][q].
- Iterate from p = 0 to p = sum1 (where sum1 is the initial sum of values of array A). We know that targetSum = sum1 - p + q, then q will be targetSum - sum1 + p.
- Answer will be minimum of all dp1[N][p] + dp2[M][q] from p = 0 to p = sum1.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to create dp1 and dp2, where
// dp[i][j] representsminimum number of
// elements required to make sum j till i-1
vector<vector<int> > minSizeSum(int a[], int n)
{
// Calculating initial sum of array
int sum = 0;
for (int i = 0; i < n; i++)
sum += a[i];
// Initialising dp with 1e9
vector<vector<int> > dp(n + 1,
vector<int>(sum + 1, 1e9));
// 0 elements are needed to make sum 0
for (int i = 0; i <= n; i++)
dp[i][0] = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
// If current element is not
// selected number of elements will
// be dp[i-1][j] and If current
// element is selected number of
// elements will be dp[i-1][j-a[i-1]]+1
if (j < a[i - 1])
dp[i][j] = dp[i - 1][j];
else
dp[i][j] = min(dp[i - 1][j],
dp[i - 1][j - a[i - 1]] + 1);
}
}
return dp;
}
// Function to calculate Minimum Operations
void minOperations(int N, int M, int targetSum, int A[],
int B[])
{
// Form Dp
vector<vector<int> > dp1 = minSizeSum(A, N),
dp2 = minSizeSum(B, M);
int sum1 = 0, sum2 = 0;
// Calculating sum of elements of both arrays
for (int i = 0; i < N; i++)
sum1 += A[i];
for (int i = 0; i < M; i++)
sum2 += B[i];
int ans = 1e9;
for (int p = 0; p <= sum1; p++) {
// targetSum = sum1-p+q
// We calculate q from p
int q = targetSum - sum1 + p;
if (q < 0 || q > sum2)
continue;
// Number of operations
ans = min(ans, dp1[N][p] + dp2[M][q]);
}
if (ans == 1e9)
ans = -1;
// Print Minimum operation Required
cout << ans << endl;
}
// Driver code
int main()
{
int N, M, targetSum = 12;
int A[] = { 1, 2, 1, 4 };
int B[] = { 5, 12, 11 };
N = sizeof(A) / sizeof(A[0]);
M = sizeof(B) / sizeof(B[0]);
// Function call
minOperations(N, M, targetSum, A, B);
return 0;
}
// Java code to implement the approach
import java.io.*;
class GFG {
// Function to create dp1 and dp2, where
// dp[i][j] representsminimum number of
// elements required to make sum j till i-1
static int[][] minSizeSum(int[] a, int n)
{
// Calculating initial sum of array
int sum = 0;
for (int i = 0; i < n; i++) {
sum += a[i];
}
// Initialising dp with 1e9
int[][] dp = new int[n + 1][sum + 1];
for (int i = 0; i < n + 1; i++) {
for (int j = 0; j < sum + 1; j++) {
dp[i][j] = (int)1e9;
}
}
// 0 elements are needed to make sum 0
for (int i = 0; i <= n; i++) {
dp[i][0] = 0;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
// If current element is not
// selected number of elements will
// be dp[i-1][j] and If current
// element is selected number of
// elements will be dp[i-1][j-a[i-1]]+1
if (j < a[i - 1]) {
dp[i][j] = dp[i - 1][j];
}
else {
dp[i][j] = Math.min(
dp[i - 1][j],
dp[i - 1][j - a[i - 1]] + 1);
}
}
}
return dp;
}
// Function to calculate Minimum Operations
static void minOperations(int N, int M, int targetSum,
int[] A, int[] B)
{
// Form Dp
int[][] dp1 = minSizeSum(A, N);
int[][] dp2 = minSizeSum(B, M);
int sum1 = 0, sum2 = 0;
// Calculating sum of elements of both arrays
for (int i = 0; i < N; i++) {
sum1 += A[i];
}
for (int i = 0; i < M; i++) {
sum2 += B[i];
}
int ans = (int)1e9;
for (int p = 0; p <= sum1; p++) {
// targetSum = sum1-p+q
// We calculate q from p
int q = targetSum - sum1 + p;
if (q < 0 || q > sum2) {
continue;
}
// Number of operations
ans = Math.min(ans, dp1[N][p] + dp2[M][q]);
}
if (ans == (int)1e9) {
ans = -1;
}
// Print Minimum operation Required
System.out.println(ans);
}
public static void main(String[] args)
{
int targetSum = 12;
int[] A = { 1, 2, 1, 4 };
int[] B = { 5, 12, 11 };
int N = A.length;
int M = B.length;
// Function call
minOperations(N, M, targetSum, A, B);
}
}
// This code is contributed by lokeshmvs21.
# Python code to implement the approach
# Function to create dp1 and dp2, where
# dp[i][j] representsminimum number of
# elements required to make sum j till i-1
def minSizeSum(a, n):
# Calculating initial sum of array
sum = 0
for i in range(n):
sum += a[i]
# Initialising dp with 1e9
dp = [[1e9 for i in range(sum + 1)]
for j in range(n + 1)]
# 0 elements are needed to make sum 0
for i in range(n + 1):
dp[i][0] = 0
for i in range(1, n + 1):
for j in range(1, sum + 1):
# If current element is not
# selected number of elements will
# be dp[i-1][j] and If current
# element is selected number of
# elements will be dp[i-1][j-a[i-1]]+1
if (j < a[i - 1]):
dp[i][j] = dp[i - 1][j]
else:
dp[i][j] = min(dp[i - 1][j],
dp[i - 1][j - a[i - 1]] + 1)
return dp
# Function to calculate Minimum Operations
def minOperations(N, M, targetSum, A, B):
# Form Dp
dp1 = minSizeSum(A, N)
dp2 = minSizeSum(B, M)
sum1 = 0
sum2 = 0
# Calculating sum of elements of both arrays
for i in range(N):
sum1 += A[i]
for i in range(M):
sum2 += B[i]
ans = 1e9
for p in range(sum1 + 1):
# targetSum = sum1-p+q
# We calculate q from p
q = targetSum - sum1 + p
if (q < 0 or q > sum2):
continue
# Number of operations
ans = min(ans, dp1[N][p] + dp2[M][q])
if (ans == 1e9):
ans = -1
# Print Minimum operation Required
print(ans)
# Driver code
if __name__ == '__main__':
targetSum = 12
A = [1, 2, 1, 4]
B = [5, 12, 11]
N = len(A)
M = len(B)
# Function call
minOperations(N, M, targetSum, A, B)
# This code is contributed by Tapesh(tapeshdua420)
// C# code to implement the approach
using System;
class GFG {
// Function to create dp1 and dp2, where
// dp[i][j] representsminimum number of
// elements required to make sum j till i-1
static int[, ] minSizeSum(int[] a, int n)
{
// Calculating initial sum of array
int sum = 0;
for (int i = 0; i < n; i++) {
sum += a[i];
}
// Initialising dp with 1e9
int[, ] dp = new int[n + 1, sum + 1];
for (int i = 0; i < n + 1; i++) {
for (int j = 0; j < sum + 1; j++) {
dp[i, j] = (int)1e9;
}
}
// 0 elements are needed to make sum 0
for (int i = 0; i <= n; i++) {
dp[i, 0] = 0;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
// If current element is not
// selected number of elements will
// be dp[i-1][j] and If current
// element is selected number of
// elements will be dp[i-1][j-a[i-1]]+1
if (j < a[i - 1]) {
dp[i, j] = dp[i - 1, j];
}
else {
dp[i, j] = Math.Min(
dp[i - 1, j],
dp[i - 1, j - a[i - 1]] + 1);
}
}
}
return dp;
}
// Function to calculate Minimum Operations
static void minOperations(int N, int M, int targetSum,
int[] A, int[] B)
{
// Form Dp
int[, ] dp1 = minSizeSum(A, N);
int[, ] dp2 = minSizeSum(B, M);
int sum1 = 0, sum2 = 0;
// Calculating sum of elements of both arrays
for (int i = 0; i < N; i++) {
sum1 += A[i];
}
for (int i = 0; i < M; i++) {
sum2 += B[i];
}
int ans = (int)1e9;
for (int p = 0; p <= sum1; p++) {
// targetSum = sum1-p+q
// We calculate q from p
int q = targetSum - sum1 + p;
if (q < 0 || q > sum2) {
continue;
}
// Number of operations
ans = Math.Min(ans, dp1[N, p] + dp2[M, q]);
}
if (ans == (int)1e9) {
ans = -1;
}
// Print Minimum operation Required
Console.WriteLine(ans);
}
public static void Main()
{
int targetSum = 12;
int[] A = { 1, 2, 1, 4 };
int[] B = { 5, 12, 11 };
int N = A.Length;
int M = B.Length;
// Function call
minOperations(N, M, targetSum, A, B);
}
}
// This code is contributed by Samim Hossain Mondal.
// JavaScript code to implement the approach
// Function to create dp1 and dp2, where
// dp[i][j] representsminimum number of
// elements required to make sum j till i-1
const minSizeSum = (a, n) => {
// Calculating initial sum of array
let sum = 0;
for (let i = 0; i < n; i++)
sum += a[i];
// Initialising dp with 1e9
let dp = new Array(n + 1).fill(1e9).map(() => new Array(sum + 1).fill(1e9));
// 0 elements are needed to make sum 0
for (let i = 0; i <= n; i++)
dp[i][0] = 0;
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= sum; j++) {
// If current element is not
// selected number of elements will
// be dp[i-1][j] and If current
// element is selected number of
// elements will be dp[i-1][j-a[i-1]]+1
if (j < a[i - 1])
dp[i][j] = dp[i - 1][j];
else
dp[i][j] = Math.min(dp[i - 1][j],
dp[i - 1][j - a[i - 1]] + 1);
}
}
return dp;
}
// Function to calculate Minimum Operations
const minOperations = (N, M, targetSum, A, B) => {
// Form Dp
let dp1 = minSizeSum(A, N), dp2 = minSizeSum(B, M);
let sum1 = 0, sum2 = 0;
// Calculating sum of elements of both arrays
for (let i = 0; i < N; i++)
sum1 += A[i];
for (let i = 0; i < M; i++)
sum2 += B[i];
let ans = 1e9;
for (let p = 0; p <= sum1; p++) {
// targetSum = sum1-p+q
// We calculate q from p
let q = targetSum - sum1 + p;
if (q < 0 || q > sum2)
continue;
// Number of operations
ans = Math.min(ans, dp1[N][p] + dp2[M][q]);
}
if (ans == 1e9)
ans = -1;
// Print Minimum operation Required
console.log(`${ans}<br/>`);
}
// Driver code
let N, M, targetSum = 12;
let A = [1, 2, 1, 4];
let B = [5, 12, 11];
N = A.length;
M = B.length;
// Function call
minOperations(N, M, targetSum, A, B);
// This code is contributed by rakeshsahni
Time Complexity: O(N * sum)
Auxiliary Space: O(N * sum)
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size sum+1.
- Set a base case by initializing the values of DP.
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- Initialize a variable ans to store the final answer and update it by iterating through the Dp.
- At last return and print the final answer stored in ans if exists else return -1.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to create dp, where
// dp[j] represents the minimum number of
// elements required to make sum j till i-1
vector<int> minSizeSum(int a[], int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += a[i];
vector<int> dp(sum + 1, 1e9);
dp[0] = 0;
for (int i = 0; i < n; i++) {
for (int j = sum; j >= a[i]; j--) {
dp[j] = min(dp[j], dp[j - a[i]] + 1);
}
}
return dp;
}
// Function to calculate Minimum Operations
void minOperations(int N, int M, int targetSum, int A[],
int B[])
{
// Form Dp
int sum_a = accumulate(A, A + N, 0);
int sum_b = accumulate(B, B + M, 0);
vector<int> dp1 = minSizeSum(A, N),
dp2 = minSizeSum(B, M);
int ans = 1e9;
for (int p = 0; p <= sum_a; p++) {
int q = targetSum - sum_a + p;
if (q < 0 || q > sum_b) {
continue;
}
// Number of operations
ans = min(ans, dp1[p] + dp2[q]);
}
if (ans == 1e9)
ans = -1;
cout << ans << endl;
}
// Driver code
int main()
{
int N, M, targetSum = 12;
int A[] = { 1, 2, 1, 4 };
int B[] = { 5, 12, 11 };
N = sizeof(A) / sizeof(A[0]);
M = sizeof(B) / sizeof(B[0]);
// Function call
minOperations(N, M, targetSum, A, B);
return 0;
}
import java.util.Arrays;
// Nikunj Sonigara
public class GFG {
static int[] minSizeSum(int[] a) {
int sum = 0;
for (int num : a)
sum += num;
int[] dp = new int[sum + 1];
Arrays.fill(dp, 1000000000);
dp[0] = 0;
for (int num : a) {
for (int j = sum; j >= num; j--) {
dp[j] = Math.min(dp[j], dp[j - num] + 1);
}
}
return dp;
}
static void minOperations(int N, int M, int targetSum, int[] A, int[] B) {
int sum_a = Arrays.stream(A).sum();
int sum_b = Arrays.stream(B).sum();
int[] dp1 = minSizeSum(A);
int[] dp2 = minSizeSum(B);
int ans = 1000000000;
for (int p = 0; p <= sum_a; p++) {
int q = targetSum - sum_a + p;
if (q < 0 || q > sum_b) {
continue;
}
ans = Math.min(ans, dp1[p] + dp2[q]);
}
if (ans == 1000000000)
ans = -1;
System.out.println(ans);
}
public static void main(String[] args) {
int N, M, targetSum = 12;
int[] A = { 1, 2, 1, 4 };
int[] B = { 5, 12, 11 };
N = A.length;
M = B.length;
minOperations(N, M, targetSum, A, B);
}
}
from typing import List
import sys
# Function to create dp, where
# dp[j] represents the minimum number of
# elements required to make sum j till i-1
def minSizeSum(a: List[int]) -> List[int]:
n = len(a)
_sum = sum(a)
dp = [sys.maxsize] * (_sum + 1)
dp[0] = 0
for i in range(n):
for j in range(_sum, a[i] - 1, -1):
dp[j] = min(dp[j], dp[j - a[i]] + 1)
return dp
# Function to calculate Minimum Operations
def minOperations(N: int, M: int, targetSum: int, A: List[int], B: List[int]) -> None:
# Form Dp
sum_a = sum(A)
sum_b = sum(B)
dp1 = minSizeSum(A)
dp2 = minSizeSum(B)
ans = sys.maxsize
for p in range(sum_a + 1):
q = targetSum - sum_a + p
if q < 0 or q > sum_b:
continue
# Number of operations
ans = min(ans, dp1[p] + dp2[q])
if ans == sys.maxsize:
ans = -1
print(ans)
# test case
if __name__ == "__main__":
targetSum = 12
A = [1, 2, 1, 4]
B = [5, 12, 11]
N = len(A)
M = len(B)
minOperations(N, M, targetSum, A, B)
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
// Function to create dp, where dp[j] represents the minimum number of
// elements required to make sum j till i-1
static List<int> MinSizeSum(int[] a, int n)
{
int sum = a.Sum();
List<int> dp = new List<int>(new int[sum + 1]);
for (int i = 1; i <= sum; i++)
{
dp[i] = 1000000009;
}
dp[0] = 0;
for (int i = 0; i < n; i++)
{
for (int j = sum; j >= a[i]; j--)
{
dp[j] = Math.Min(dp[j], dp[j - a[i]] + 1);
}
}
return dp;
}
// Function to calculate Minimum Operations
static void MinOperations(int N, int M, int targetSum, int[] A, int[] B)
{
// Form Dp
int sum_a = A.Sum();
int sum_b = B.Sum();
List<int> dp1 = MinSizeSum(A, N);
List<int> dp2 = MinSizeSum(B, M);
int ans = 1000000009;
for (int p = 0; p <= sum_a; p++)
{
int q = targetSum - sum_a + p;
if (q < 0 || q > sum_b)
{
continue;
}
// Number of operations
ans = Math.Min(ans, dp1[p] + dp2[q]);
}
if (ans == 1000000009)
{
ans = -1;
}
Console.WriteLine(ans);
}
// Driver code
static void Main(string[] args)
{
int N, M, targetSum = 12;
int[] A = { 1, 2, 1, 4 };
int[] B = { 5, 12, 11 };
N = A.Length;
M = B.Length;
// Function call
MinOperations(N, M, targetSum, A, B);
}
}
// Function to create dp, where
// dp[j] represents the minimum number of
// elements required to make sum j till i-1
function minSizeSum(a, n) {
let sum = 0;
for (let i = 0; i < n; i++) {
sum += a[i];
}
let dp = new Array(sum + 1).fill(1e9);
dp[0] = 0;
for (let i = 0; i < n; i++) {
for (let j = sum; j >= a[i]; j--) {
dp[j] = Math.min(dp[j], dp[j - a[i]] + 1);
}
}
return dp;
}
// Function to calculate Minimum Operations
function minOperations(N, M, targetSum, A, B) {
// Form Dp
let sum_a = A.reduce((acc, curr) => acc + curr, 0);
let sum_b = B.reduce((acc, curr) => acc + curr, 0);
let dp1 = minSizeSum(A, N);
let dp2 = minSizeSum(B, M);
let ans = 1e9;
for (let p = 0; p <= sum_a; p++) {
let q = targetSum - sum_a + p;
if (q < 0 || q > sum_b) {
continue;
}
// Number of operations
ans = Math.min(ans, dp1[p] + dp2[q]);
}
if (ans == 1e9) {
ans = -1;
}
console.log(ans);
}
// Test case
let N, M, targetSum = 12;
let A = [1, 2, 1, 4];
let B = [5, 12, 11];
N = A.length;
M = B.length;
minOperations(N, M, targetSum, A, B);
Time Complexity: O(N * sum)
Auxiliary Space: O(sum)
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Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
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Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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