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Maximize array sum by concatenating corresponding elements of given two arrays
  • Last Updated : 13 Apr, 2021

Given two array A[] and B[] of the same length, the task is to find the maximum array sum that can be formed by joining the corresponding elements of the array in any order.
 

Input: A[] = {1, 2, 3, 4, 5}, B[] = {3, 2, 1, 4, 5} 
Output: 183 
Explanation: 
Numbers formed by joining the digits of the elements are – 
Join(A[0], B[0]) = Join(1, 3) = 13 or 31 
Join(A[1], B[1]) = Join(2, 2) = 22 
Join(A[2], B[2]) = Join(3, 1) = 31 or 13 
Join(A[3], B[3]) = Join(4, 4) = 44 
Join(A[4], B[4]) = Join(5, 5) = 55 
Therefore for maximum sum, the array will be {31, 22, 31, 44, 55} with sum = 183
Input: A[] = {11, 23, 38, 43, 59}, B[] = {36, 24, 17, 40, 56} 
Output: 19850 
Explanation: 
Numbers formed by joining the digits of the elements are – 
Join(A[0], B[0]) = Join(11, 36) = 1136 or 3611 
Join(A[1], B[1]) = Join(23, 24) = 2324 or 2423 
Join(A[2], B[2]) = Join(38, 17) = 3817 or 1738 
Join(A[3], B[3]) = Join(43, 40) = 4340 or 4043 
Join(A[4], B[4]) = Join(59, 56) = 5956 or 5659 
Therefore for maximum sum, the array will be {3611, 2423, 3817, 4340, 5956} with sum = 19850 
 

 

Approach: The idea is to iterate over the array and for each corresponding element of the array 
 

  1. join them together by iterating over the digits of one number and add the digits into another number which can be defined as follows: 
     
// Join the numbers
for digits in numberA:
    numberB = numberB*10 + digit
  1.  
  2. Similarly, join the numbers in reverse order and take the maximum of those numbers.
  3. Update the corresponding element of the resultant array with the maximum among the two.
  4. Similarly find all elements of the resultant array and compute the sum.

Below is the implementation of the above approach: 
 

C++




// C++ implementation to find the
// maximum array sum by concatenating
// corresponding elements of given two arrays
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to join the two numbers
int joinNumbers(int numA, int numB)
{
    int revB = 0;
 
    // Loop to reverse the digits
    // of the one number
    while (numB > 0) {
        revB = revB * 10 + (numB % 10);
        numB = numB / 10;
    }
 
    // Loop to join two numbers
    while (revB > 0) {
        numA = numA * 10 + (revB % 10);
        revB = revB / 10;
    }
    return numA;
}
 
// Function to find the maximum array sum
int findMaxSum(int A[], int B[], int n)
{
    int maxArr[n];
 
    // Loop to iterate over the two
    // elements of the array
    for (int i = 0; i < n; ++i) {
 
        int X = joinNumbers(A[i], B[i]);
        int Y = joinNumbers(B[i], A[i]);
        int mx = max(X, Y);
        maxArr[i] = mx;
    }
 
    // Find the array sum
    int maxAns = 0;
    for (int i = 0; i < n; i++) {
        maxAns += maxArr[i];
    }
 
    // Return the array sum
    return maxAns;
}
 
// Driver Code
int main()
{
    int N = 5;
    int A[5] = { 11, 23, 38, 43, 59 };
    int B[5] = { 36, 24, 17, 40, 56 };
 
    cout << findMaxSum(A, B, N);
}

Java




// Java implementation to find the
// maximum array sum by concatenating
// corresponding elements of given two arrays
class GFG {
 
// Function to join the two numbers
static int joinNumbers(int numA, int numB)
{
    int revB = 0;
 
    // Loop to reverse the digits
    // of the one number
    while (numB > 0)
    {
        revB = revB * 10 + (numB % 10);
        numB = numB / 10;
    }
 
    // Loop to join two numbers
    while (revB > 0)
    {
        numA = numA * 10 + (revB % 10);
        revB = revB / 10;
    }
 
    return numA;
}
 
// Function to find the maximum array sum
static int findMaxSum(int A[], int B[], int n)
{
    int maxArr[] = new int[n];
 
    // Loop to iterate over the two
    // elements of the array
    for(int i = 0; i < n; ++i)
    {
       int X = joinNumbers(A[i], B[i]);
       int Y = joinNumbers(B[i], A[i]);
       int mx = Math.max(X, Y);
 
       maxArr[i] = mx;
    }
 
    // Find the array sum
    int maxAns = 0;
    for(int i = 0; i < n; i++)
    {
       maxAns += maxArr[i];
    }
 
    // Return the array sum
    return maxAns;
}
 
// Driver Code
public static void main(String args[])
{
    int N = 5;
    int A[] = { 11, 23, 38, 43, 59 };
    int B[] = { 36, 24, 17, 40, 56 };
 
    System.out.println(findMaxSum(A, B, N));
}
}
 
// This code is contributed by rutvik_56   

Python3




# Python3 implementation to find the
# maximum array sum by concatenating
# corresponding elements of given two arrays
 
# Function to join the two numbers
def joinNumbers(numA, numB):
    revB = 0
 
    # Loop to reverse the digits
    # of the one number
    while (numB > 0):
        revB = revB * 10 + (numB % 10)
        numB = numB // 10
 
    # Loop to join two numbers
    while (revB > 0):
        numA = numA * 10 + (revB % 10)
        revB = revB // 10
 
    return numA
 
# Function to find the maximum array sum
def findMaxSum(A, B, n):
    maxArr = [0 for i in range(n)]
 
    # Loop to iterate over the two
    # elements of the array
    for i in range(n):
         
        X = joinNumbers(A[i], B[i])
        Y = joinNumbers(B[i], A[i])
        mx = max(X, Y)
        maxArr[i] = mx
 
    # Find the array sum
    maxAns = 0
     
    for i in range(n):
        maxAns += maxArr[i]
 
    # Return the array sum
    return maxAns
 
# Driver Code
if __name__ == '__main__':
     
    N = 5
    A = [ 11, 23, 38, 43, 59 ]
    B = [ 36, 24, 17, 40, 56 ]
 
    print(findMaxSum(A, B, N))
 
# This code is contributed by Samarth

C#




// C# implementation to find the
// maximum array sum by concatenating
// corresponding elements of given two arrays
using System;
class GFG{
 
// Function to join the two numbers
static int joinNumbers(int numA, int numB)
{
    int revB = 0;
 
    // Loop to reverse the digits
    // of the one number
    while (numB > 0)
    {
        revB = revB * 10 + (numB % 10);
        numB = numB / 10;
    }
 
    // Loop to join two numbers
    while (revB > 0)
    {
        numA = numA * 10 + (revB % 10);
        revB = revB / 10;
    }
 
    return numA;
}
 
// Function to find the maximum array sum
static int findMaxSum(int []A, int []B, int n)
{
    int []maxArr = new int[n];
 
    // Loop to iterate over the two
    // elements of the array
    for(int i = 0; i < n; ++i)
    {
        int X = joinNumbers(A[i], B[i]);
        int Y = joinNumbers(B[i], A[i]);
        int mx = Math.Max(X, Y);
     
        maxArr[i] = mx;
    }
 
    // Find the array sum
    int maxAns = 0;
    for(int i = 0; i < n; i++)
    {
        maxAns += maxArr[i];
    }
 
    // Return the array sum
    return maxAns;
}
 
// Driver Code
public static void Main(String []args)
{
    int N = 5;
    int []A = { 11, 23, 38, 43, 59 };
    int []B = { 36, 24, 17, 40, 56 };
 
    Console.WriteLine(findMaxSum(A, B, N));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation to find the
// maximum array sum by concatenating
// corresponding elements of given two arrays
 
// Function to join the two numbers
function joinNumbers(numA, numB)
{
    var revB = 0;
 
    // Loop to reverse the digits
    // of the one number
    while (numB > 0) {
        revB = revB * 10 + (numB % 10);
        numB = parseInt(numB / 10);
    }
 
    // Loop to join two numbers
    while (revB > 0) {
        numA = numA * 10 + (revB % 10);
        revB = parseInt( revB / 10);
    }
    return numA;
}
 
// Function to find the maximum array sum
function findMaxSum(A, B, n)
{
    var maxArr = Array(n).fill(0);
 
    // Loop to iterate over the two
    // elements of the array
    for (var i = 0; i < n; ++i) {
 
        var X = joinNumbers(A[i], B[i]);
        var Y = joinNumbers(B[i], A[i]);
        var mx = Math.max(X, Y);
        maxArr[i] = mx;
    }
 
    // Find the array sum
    var maxAns = 0;
    for (var i = 0; i < n; i++) {
        maxAns += maxArr[i];
    }
 
    // Return the array sum
    return maxAns;
}
 
// Driver Code
var N = 5;
var A = [ 11, 23, 38, 43, 59 ];
var B = [ 36, 24, 17, 40, 56 ];
document.write( findMaxSum(A, B, N));
 
</script>
Output: 
19850

 

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