Minimum sum of two elements from two arrays such that indexes are not same
Last Updated :
23 Jul, 2025
Given two arrays a[] and b[] of same size. Task is to find minimum sum of two elements such that they belong to different arrays and are not at same index in their arrays.
Examples:
Input : a[] = {5, 4, 13, 2, 1}
b[] = {2, 3, 4, 6, 5}
Output : 3
We take 1 from a[] and 2 from b[]
Sum is 1 + 2 = 3.
Input : a[] = {5, 4, 13, 1}
b[] = {3, 2, 6, 1}
Output : 3
We take 1 from a[] and 2 from b[].
Note that we can't take 1 from b[]
as the elements can not be at same
index. A simple solution is to consider every element of a[], form its pair with all elements of b[] at indexes different from its index and compute sums. Finally return the minimum sum. Time complexity of this solution is O(n2)
An efficient solution works in O(n) time. Below are steps.
- Find minimum elements from a[] and b[]. Let these elements be minA and minB respectively.
- If indexes of minA and minB are not same, return minA + minB.
- Else find second minimum elements from two arrays. Let these elements be minA2 and minB2. Return min(minA + minB2, minA2 + minB)
Below is the implementation of above idea:
C++
// C++ program to find minimum sum of two
// elements chosen from two arrays such that
// they are not at same index.
#include <bits/stdc++.h>
using namespace std;
// Function which returns minimum sum of two
// array elements such that their indexes are
// not same
int minSum(int a[], int b[], int n)
{
// Finding minimum element in array A and
// also/ storing its index value.
int minA = a[0], indexA;
for (int i=1; i<n; i++)
{
if (a[i] < minA)
{
minA = a[i];
indexA = i;
}
}
// Finding minimum element in array B and
// also storing its index value
int minB = b[0], indexB;
for (int i=1; i<n; i++)
{
if (b[i] < minB)
{
minB = b[i];
indexB = i;
}
}
// If indexes of minimum elements are
// not same, return their sum.
if (indexA != indexB)
return (minA + minB);
// When index of A is not same as previous
// and value is also less than other minimum
// Store new minimum and store its index
int minA2 = INT_MAX, indexA2;
for (int i=0; i<n; i++)
{
if (i != indexA && a[i] < minA2)
{
minA2 = a[i];
indexA2 = i;
}
}
// When index of B is not same as previous
// and value is also less than other minimum.
// Store new minimum and store its index
int minB2 = INT_MAX, indexB2;
for (int i=0; i<n; i++)
{
if (i != indexB && b[i] < minB2)
{
minB2 = b[i];
indexB2 = i;
}
}
// Taking sum of previous minimum of a[]
// with new minimum of b[]
// and also sum of previous minimum of b[]
// with new minimum of a[]
// and return whichever is minimum.
return min(minB + minA2, minA + minB2);
}
// Driver code
int main()
{
int a[] = {5, 4, 3, 8, 1};
int b[] = {2, 3, 4, 2, 1};
int n = sizeof(a)/sizeof(a[0]);
cout << minSum(a, b, n);
return 0;
}
// Java program to find minimum sum of two
// elements chosen from two arrays such that
// they are not at same index.
class Minimum{
// Function which returns minimum sum of two
// array elements such that their indexes are
// not same
public static int minSum(int a[], int b[], int n)
{
// Finding minimum element in array A and
// also/ storing its index value.
int minA = a[0], indexA = 0;
for (int i=1; i<n; i++)
{
if (a[i] < minA)
{
minA = a[i];
indexA = i;
}
}
// Finding minimum element in array B and
// also storing its index value
int minB = b[0], indexB = 0;
for (int i=1; i<n; i++)
{
if (b[i] < minB)
{
minB = b[i];
indexB = i;
}
}
// If indexes of minimum elements are
// not same, return their sum.
if (indexA != indexB)
return (minA + minB);
// When index of A is not same as previous
// and value is also less than other minimum
// Store new minimum and store its index
int minA2 = Integer.MAX_VALUE, indexA2 = 0;
for (int i=0; i<n; i++)
{
if (i != indexA && a[i] < minA2)
{
minA2 = a[i];
indexA2 = i;
}
}
// When index of B is not same as previous
// and value is also less than other minimum.
// Store new minimum and store its index
int minB2 = Integer.MAX_VALUE, indexB2 = 0;
for (int i=0; i<n; i++)
{
if (i != indexB && b[i] < minB2)
{
minB2 = b[i];
indexB2 = i;
}
}
// Taking sum of previous minimum of a[]
// with new minimum of b[]
// and also sum of previous minimum of b[]
// with new minimum of a[]
// and return whichever is minimum.
return Math.min(minB + minA2, minA + minB2);
}
public static void main(String[] args)
{
int a[] = {5, 4, 3, 8, 1};
int b[] = {2, 3, 4, 2, 1};
int n = 5;
System.out.print(minSum(a, b, n));
}
}
// This code is contributed by rishabh_jain
# Python3 code to find minimum sum of
# two elements chosen from two arrays
# such that they are not at same index.
import sys
# Function which returns minimum sum
# of two array elements such that their
# indexes arenot same
def minSum(a, b, n):
# Finding minimum element in array A
# and also storing its index value.
minA = a[0]
indexA = 0
for i in range(1,n):
if a[i] < minA:
minA = a[i]
indexA = i
# Finding minimum element in array B
# and also storing its index value
minB = b[0]
indexB = 0
for i in range(1, n):
if b[i] < minB:
minB = b[i]
indexB = i
# If indexes of minimum elements
# are not same, return their sum.
if indexA != indexB:
return (minA + minB)
# When index of A is not same as
# previous and value is also less
# than other minimum. Store new
# minimum and store its index
minA2 = sys.maxsize
indexA2=0
for i in range(n):
if i != indexA and a[i] < minA2:
minA2 = a[i]
indexA2 = i
# When index of B is not same as
# previous and value is also less
# than other minimum. Store new
# minimum and store its index
minB2 = sys.maxsize
indexB2 = 0
for i in range(n):
if i != indexB and b[i] < minB2:
minB2 = b[i]
indexB2 = i
# Taking sum of previous minimum of
# a[] with new minimum of b[]
# and also sum of previous minimum
# of b[] with new minimum of a[]
# and return whichever is minimum.
return min(minB + minA2, minA + minB2)
# Driver code
a = [5, 4, 3, 8, 1]
b = [2, 3, 4, 2, 1]
n = len(a)
print(minSum(a, b, n))
# This code is contributed by "Sharad_Bhardwaj".
// C# program to find minimum sum of
// two elements chosen from two arrays
// such that they are not at same index.
using System;
public class GFG {
// Function which returns minimum
// sum of two array elements such
// that their indexes are not same
static int minSum(int []a, int []b,
int n)
{
// Finding minimum element in
// array A and also/ storing its
// index value.
int minA = a[0], indexA = 0;
for (int i = 1; i < n; i++)
{
if (a[i] < minA)
{
minA = a[i];
indexA = i;
}
}
// Finding minimum element in
// array B and also storing its
// index value
int minB = b[0], indexB = 0;
for (int i = 1; i < n; i++)
{
if (b[i] < minB)
{
minB = b[i];
indexB = i;
}
}
// If indexes of minimum elements
// are not same, return their sum.
if (indexA != indexB)
return (minA + minB);
// When index of A is not same as
// previous and value is also less
// than other minimum Store new
// minimum and store its index
int minA2 = int.MaxValue;
for (int i=0; i<n; i++)
{
if (i != indexA && a[i] < minA2)
{
minA2 = a[i];
}
}
// When index of B is not same as
// previous and value is also less
// than other minimum. Store new
// minimum and store its index
int minB2 = int.MaxValue;
for (int i=0; i<n; i++)
if (i != indexB && b[i] < minB2)
minB2 = b[i];
// Taking sum of previous minimum
// of a[] with new minimum of b[]
// and also sum of previous minimum
// of b[] with new minimum of a[]
// and return whichever is minimum.
return Math.Min(minB + minA2,
minA + minB2);
}
public static void Main()
{
int []a = {5, 4, 3, 8, 1};
int []b = {2, 3, 4, 2, 1};
int n = 5;
Console.Write(minSum(a, b, n));
}
}
// This code is contributed by Sam007.
<script>
// JavaScript program to find minimum sum of two
// elements chosen from two arrays such that
// they are not at same index.
// Function which returns minimum sum of two
// array elements such that their indexes are
// not same
function minSum(a, b, n)
{
// Finding minimum element in array A and
// also/ storing its index value.
let minA = a[0], indexA;
for (let i=1; i<n; i++)
{
if (a[i] < minA)
{
minA = a[i];
indexA = i;
}
}
// Finding minimum element in array B and
// also storing its index value
let minB = b[0], indexB;
for (let i=1; i<n; i++)
{
if (b[i] < minB)
{
minB = b[i];
indexB = i;
}
}
// If indexes of minimum elements are
// not same, return their sum.
if (indexA != indexB)
return (minA + minB);
// When index of A is not same as previous
// and value is also less than other minimum
// Store new minimum and store its index
let minA2 = Number.MAX_SAFE_INTEGER, indexA2;
for (let i=0; i<n; i++)
{
if (i != indexA && a[i] < minA2)
{
minA2 = a[i];
indexA2 = i;
}
}
// When index of B is not same as previous
// and value is also less than other minimum.
// Store new minimum and store its index
let minB2 = Number.MAX_SAFE_INTEGER, indexB2;
for (let i=0; i<n; i++)
{
if (i != indexB && b[i] < minB2)
{
minB2 = b[i];
indexB2 = i;
}
}
// Taking sum of previous minimum of a[]
// with new minimum of b[]
// and also sum of previous minimum of b[]
// with new minimum of a[]
// and return whichever is minimum.
return Math.min(minB + minA2, minA + minB2);
}
// Driver code
let a = [5, 4, 3, 8, 1];
let b = [2, 3, 4, 2, 1];
let n = a.length;
document.write(minSum(a, b, n));
// This code is contributed by Surbhi Tyagi.
</script>
<?php
// PHP program to find minimum
// sum of two elements chosen
// from two arrays such that
// they are not at same index.
// Function which returns
// minimum sum of two array
// elements such that their
// indexes are not same
function minSum($a, $b, $n)
{
// Finding minimum element
// in array A and also
// storing its index value.
$minA = $a[0];
for ($i = 1; $i < $n; $i++)
{
if ($a[$i] < $minA)
{
$minA = $a[$i];
$indexA = $i;
}
}
// Finding minimum element
// in array B and also
// storing its index value
$minB = $b[0];
for ($i = 1; $i < $n; $i++)
{
if ($b[$i] < $minB)
{
$minB = $b[$i];
$indexB = $i;
}
}
// If indexes of minimum
// elements are not same,
// return their sum.
if ($indexA != $indexB)
return ($minA + $minB);
// When index of A is not
// same as previous and
// value is also less than
// other minimum. Store new
// minimum and store its index
$minA2 = 9999999;
$indexA2 = 0;
for ($i = 0; $i < $n; $i++)
{
if ($i != $indexA &&
$a[$i] < $minA2)
{
$minA2 = $a[$i];
$indexA2 = $i;
}
}
// When index of B is not
// same as previous and
// value is also less than
// other minimum. Store new
// minimum and store its index
$minB2 = 999999;
$indexB2 = 0;
for ($i = 0; $i < $n; $i++)
{
if ($i != $indexB &&
$b[$i] < $minB2)
{
$minB2 = $b[$i];
$indexB2 = $i;
}
}
// Taking sum of previous
// minimum of a[] with
// new minimum of b[]
// and also sum of previous
// minimum of b[] with new
// minimum of a[]
// and return whichever
// is minimum.
return min($minB + $minA2,
$minA + $minB2);
}
// Driver code
$a = array(5, 4, 3, 8, 1);
$b = array(2, 3, 4, 2, 1);
$n = count($a);
echo minSum($a, $b, $n);
// This code is contributed
// by Sam007
?>
Time Complexity : O(n)
Auxiliary Space : O(1)
New approach:- Here , Another approach to solve this problem is by using sorting. We can sort both arrays in non-decreasing order and then find the minimum sum by taking the sum of the first two elements of the sorted arrays.
Algorithm:
- Define a function minSum which takes input arrays a, b and their length n as arguments.
- Sort both arrays a and b in non-decreasing order using the Arrays.sort method.
- Initialize a variable minSum to Integer.MAX_VALUE.
- Initialize two variables i and j to 0.
- Use a while loop to iterate over the arrays a and b until either i or j is less than n.
- Check if the indexes of the current elements being compared are not the same, calculate their sum and if it is less than the current minSum, update the minSum.
- If the element in a at index i is less than the element in b at index j, increment i by 1. Otherwise, increment j by 1.
- Return the minimum sum calculated in step 3.
- In the main method, define two arrays a and b, set their values and their length n.
- Call the minSum function with arrays a, b and n as arguments and print the result.
Here is the implementation of this approach:-
C++
#include <bits/stdc++.h>
#include <algorithm>
#include <vector>
using namespace std;
int minSum(vector<int>& a, vector<int>& b, int n) {
// Sort both arrays in non-decreasing order
sort(a.begin(), a.end());
sort(b.begin(), b.end());
// Initialize the minimum sum
int minSum = INT_MAX;
// Iterate over the arrays and find the minimum sum
int i = 0, j = 0;
while (i < n && j < n) {
if (i != j) {
int sum = a[i] + b[j];
if (sum < minSum) {
minSum = sum;
}
}
if (a[i] < b[j]) {
i++;
} else {
j++;
}
}
// Return the minimum sum
return minSum;
}
int main() {
vector<int> a = {5, 4, 3, 8, 1};
vector<int> b = {2, 3, 4, 2, 1};
int n = 5;
cout << minSum(a, b, n) << endl;
return 0;
}
import java.util.Arrays;
class Minimum {
public static int minSum(int a[], int b[], int n) {
// Sort both arrays in non-decreasing order
Arrays.sort(a);
Arrays.sort(b);
// Initialize the minimum sum
int minSum = Integer.MAX_VALUE;
// Iterate over the arrays and find the minimum sum
int i = 0, j = 0;
while (i < n && j < n) {
if (i != j) {
int sum = a[i] + b[j];
if (sum < minSum) {
minSum = sum;
}
}
if (a[i] < b[j]) {
i++;
} else {
j++;
}
}
// Return the minimum sum
return minSum;
}
public static void main(String[] args) {
int a[] = {5, 4, 3, 8, 1};
int b[] = {2, 3, 4, 2, 1};
int n = 5;
System.out.print(minSum(a, b, n));
}
}
def minSum(a, b, n):
# Sort both arrays in non-decreasing order
a.sort()
b.sort()
# Initialize the minimum sum
minSum = float('inf')
# Iterate over the arrays and find the minimum sum
i, j = 0, 0
while i < n and j < n:
if i != j:
sum_val = a[i] + b[j]
if sum_val < minSum:
minSum = sum_val
if a[i] < b[j]:
i += 1
else:
j += 1
# Return the minimum sum
return minSum
a = [5, 4, 3, 8, 1]
b = [2, 3, 4, 2, 1]
n = 5
print(minSum(a, b, n))
using System;
using System.Collections.Generic;
using System.Linq;
class Program {
// Function to find the minimum sum of elements from two
// sorted arrays
static int MinSum(List<int> a, List<int> b, int n)
{
// Sort both arrays in non-decreasing order
a.Sort();
b.Sort();
// Initialize the minimum sum
int minSum = int.MaxValue;
// Iterate over the arrays and find the minimum sum
int i = 0, j = 0;
while (i < n && j < n) {
if (i != j) {
int sum = a[i] + b[j];
if (sum < minSum) {
minSum = sum;
}
}
if (a[i] < b[j]) {
i++;
}
else {
j++;
}
}
// Return the minimum sum
return minSum;
}
static void Main(string[] args)
{
List<int> a = new List<int>{ 5, 4, 3, 8, 1 };
List<int> b = new List<int>{ 2, 3, 4, 2, 1 };
int n = 5;
Console.WriteLine(MinSum(a, b, n));
}
}
function minSum(a, b, n) {
// Sort both arrays in non-decreasing order
a.sort((x, y) => x - y);
b.sort((x, y) => x - y);
// Initialize the minimum sum
let minSum = Number.MAX_SAFE_INTEGER;
// Iterate over the arrays and find the minimum sum
let i = 0, j = 0;
while (i < n && j < n) {
if (i !== j) {
const sum = a[i] + b[j];
if (sum < minSum) {
minSum = sum;
}
}
if (a[i] < b[j]) {
i++;
} else {
j++;
}
}
// Return the minimum sum
return minSum;
}
// Driver code
const a = [5, 4, 3, 8, 1];
const b = [2, 3, 4, 2, 1];
const n = 5;
console.log(minSum(a, b, n));
Output:-
3
Time Complexity:- The time complexity of this approach is O(n log n), dominated by the sorting of the arrays using Arrays.sort() which takes O(n log n) time complexity. The while loop then iterates over both arrays once, which takes O(n) time complexity. Therefore, the overall time complexity is O(n log n).
Auxiliary space:- The auxiliary space complexity is O(1) as we are not using any additional data structures to solve the problem, only some variables to store the minimum sum and the current indices of the arrays.
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