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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 returnss 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 returnss 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 contibuted 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 returnss 
// 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 
?> 

Output:

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

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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