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

Difficulty Level : Easy
Last Updated : 30 Jul, 2018

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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Related Articles

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

C++

Java

Python3

C#

PHP