# Minimum sum of two elements from two arrays such that indexes are not same

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(n^{2})

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

`// 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

`# 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#

`// 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.` |

## PHP

`<?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` `?>` |

## Javascript

`<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>` |

**Output:**

3

**Time Complexity :** O(n) **Auxiliary Space :** O(1)

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