Maximum Sum Path in Two Arrays - GeeksforGeeks

14 Jan, 2022
Given two sorted arrays, such that the arrays may have some common elements. Find the sum of the maximum sum path to reach from the beginning of any array to end of any of the two arrays. We can switch from one array to another array only at common elements. Note: The common elements do not have to be at the same indexes.
Expected Time Complexity: O(m+n), where m is the number of elements in ar1[] and n is the number of elements in ar2[].
Examples: 
Input: ar1[] = {2, 3, 7, 10, 12}
       ar2[] = {1, 5, 7, 8}
Output: 35

Explanation: 35 is sum of 1 + 5 + 7 + 10 + 12.
We start from the first element of arr2 which is 1, then we
move to 5, then 7.  From 7, we switch to ar1 (as 7 is common)
and traverse 10 and 12.

Input: ar1[] = {10, 12}
       ar2 = {5, 7, 9}
Output: 22

Explanation: 22 is the sum of 10 and 12.
Since there is no common element, we need to take all 
elements from the array with more sum.

Input: ar1[] = {2, 3, 7, 10, 12, 15, 30, 34}
        ar2[] = {1, 5, 7, 8, 10, 15, 16, 19}
Output: 122

Explanation: 122 is sum of 1, 5, 7, 8, 10, 12, 15, 30, 34
Efficient Approach: The idea is to do something similar to merge process of merge sort. This involves calculating the sum of elements between all common points of both arrays. Whenever there is a common point, compare the two sums and add the maximum of two to the result.
Algorithm: 
Create some variables, result, sum1, sum2. Initialize result as 0. Also initialize two variables sum1 and sum2 as 0. Here sum1 and sum2 are used to store sum of element in ar1[] and ar2[] respectively. These sums are between two common points.Now run a loop to traverse elements of both arrays. While traversing compare current elements of array 1 and array 2 in the following order.If current element of array 1 is smaller than current element of array 2, then update sum1, else if current element of array 2 is smaller, then update sum2.If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. Also add the common element to the result.This step can be compared to the merging of two sorted arrays, If the smallest element of the two current array indices is processed then it is guaranteed that if there is any common element it will be processed together. So the sum of elements between two common elements can be processed.
Create some variables, result, sum1, sum2. Initialize result as 0. Also initialize two variables sum1 and sum2 as 0. Here sum1 and sum2 are used to store sum of element in ar1[] and ar2[] respectively. These sums are between two common points.
Now run a loop to traverse elements of both arrays. While traversing compare current elements of array 1 and array 2 in the following order.If current element of array 1 is smaller than current element of array 2, then update sum1, else if current element of array 2 is smaller, then update sum2.If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. Also add the common element to the result.This step can be compared to the merging of two sorted arrays, If the smallest element of the two current array indices is processed then it is guaranteed that if there is any common element it will be processed together. So the sum of elements between two common elements can be processed.
If current element of array 1 is smaller than current element of array 2, then update sum1, else if current element of array 2 is smaller, then update sum2.If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. Also add the common element to the result.This step can be compared to the merging of two sorted arrays, If the smallest element of the two current array indices is processed then it is guaranteed that if there is any common element it will be processed together. So the sum of elements between two common elements can be processed.
If current element of array 1 is smaller than current element of array 2, then update sum1, else if current element of array 2 is smaller, then update sum2.
If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. Also add the common element to the result.
This step can be compared to the merging of two sorted arrays, If the smallest element of the two current array indices is processed then it is guaranteed that if there is any common element it will be processed together. So the sum of elements between two common elements can be processed.
Below is the implementation of the above code: 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find maximum sum path#include <iostream>using namespace std; // Utility function to find maximum of two integersint max(int x, int y) { return (x > y) ? x : y; } // This function returns the sum of elements on maximum path// from beginning to endint maxPathSum(int ar1[], int ar2[], int m, int n){    // initialize indexes for ar1[] and ar2[]    int i = 0, j = 0;     // Initialize result and current sum through ar1[] and    // ar2[].    int result = 0, sum1 = 0, sum2 = 0;     // Below 3 loops are similar to merge in merge sort    while (i < m && j < n)    {        // Add elements of ar1[] to sum1        if (ar1[i] < ar2[j])            sum1 += ar1[i++];         // Add elements of ar2[] to sum2        else if (ar1[i] > ar2[j])            sum2 += ar2[j++];         else // we reached a common point        {            // Take the maximum of two sums and add to            // result              //Also add the common element of array, once            result += max(sum1, sum2) + ar1[i];             // Update sum1 and sum2 for elements after this            // intersection point            sum1 = 0;              sum2 = 0;                         //update i and j to move to next element of each array              i++;              j++;         }    }     // Add remaining elements of ar1[]    while (i < m)        sum1 += ar1[i++];     // Add remaining elements of ar2[]    while (j < n)        sum2 += ar2[j++];     // Add maximum of two sums of remaining elements    result += max(sum1, sum2);     return result;} // Driver codeint main(){    int ar1[] = { 2, 3, 7, 10, 12, 15, 30, 34 };    int ar2[] = { 1, 5, 7, 8, 10, 15, 16, 19 };    int m = sizeof(ar1) / sizeof(ar1[0]);    int n = sizeof(ar2) / sizeof(ar2[0]);       // Function call    cout << "Maximum sum path is "         << maxPathSum(ar1, ar2, m, n);    return 0;}
// JAVA program to find maximum sum pathclass MaximumSumPath{    // Utility function to find maximum of two integers    int max(int x, int y) { return (x > y) ? x : y; }     // This function returns the sum of elements on maximum    // path from beginning to end    int maxPathSum(int ar1[], int ar2[], int m, int n)    {        // initialize indexes for ar1[] and ar2[]        int i = 0, j = 0;         // Initialize result and current sum through ar1[]        // and ar2[].        int result = 0, sum1 = 0, sum2 = 0;         // Below 3 loops are similar to merge in merge sort        while (i < m && j < n)        {            // Add elements of ar1[] to sum1            if (ar1[i] < ar2[j])                sum1 += ar1[i++];             // Add elements of ar2[] to sum2            else if (ar1[i] > ar2[j])                sum2 += ar2[j++];             // we reached a common point            else            {                // Take the maximum of two sums and add to                // result                //Also add the common element of array, once                result += max(sum1, sum2) + ar1[i];                 // Update sum1 and sum2 for elements after this                // intersection point                sum1 = 0;                  sum2 = 0;                 //update i and j to move to next element of each array                i++;                j++;            }        }         // Add remaining elements of ar1[]        while (i < m)            sum1 += ar1[i++];         // Add remaining elements of ar2[]        while (j < n)            sum2 += ar2[j++];         // Add maximum of two sums of remaining elements        result += max(sum1, sum2);         return result;    }     // Driver code    public static void main(String[] args)    {        MaximumSumPath sumpath = new MaximumSumPath();        int ar1[] = { 2, 3, 7, 10, 12, 15, 30, 34 };        int ar2[] = { 1, 5, 7, 8, 10, 15, 16, 19 };        int m = ar1.length;        int n = ar2.length;               // Function call        System.out.println(            "Maximum sum path is :"            + sumpath.maxPathSum(ar1, ar2, m, n));    }} // This code has been contributed by Mayank Jaiswal
# Python program to find maximum sum path # This function returns the sum of elements on maximum path from# beginning to end  def maxPathSum(ar1, ar2, m, n):     # initialize indexes for ar1[] and ar2[]    i, j = 0, 0     # Initialize result and current sum through ar1[] and ar2[]    result, sum1, sum2 = 0, 0, 0     # Below 3 loops are similar to merge in merge sort    while (i < m and j < n):         # Add elements of ar1[] to sum1        if ar1[i] < ar2[j]:            sum1 += ar1[i]            i += 1         # Add elements of ar2[] to sum2        elif ar1[i] > ar2[j]:            sum2 += ar2[j]            j += 1         else:   # we reached a common point             # Take the maximum of two sums and add to result            result += max(sum1, sum2) +ar1[i]            #update sum1 and sum2 to be considered fresh for next elements            sum1 = 0            sum2 = 0            #update i and j to move to next element in each array            i +=1            j +=1                 # Add remaining elements of ar1[]    while i < m:        sum1 += ar1[i]        i += 1    # Add remaining elements of b[]    while j < n:        sum2 += ar2[j]        j += 1     # Add maximum of two sums of remaining elements    result += max(sum1, sum2)     return result  # Driver codear1 = [2, 3, 7, 10, 12, 15, 30, 34]ar2 = [1, 5, 7, 8, 10, 15, 16, 19]m = len(ar1)n = len(ar2) # Function callprint ("Maximum sum path is", maxPathSum(ar1, ar2, m, n)) # This code is contributed by __Devesh Agrawal__
// C# program for Maximum Sum Path in// Two Arraysusing System; class GFG {     // Utility function to find maximum    // of two integers    static int max(int x, int y) { return (x > y) ? x : y; }     // This function returns the sum of    // elements on maximum path from    // beginning to end    static int maxPathSum(int[] ar1, int[] ar2, int m,                          int n)    {         // initialize indexes for ar1[]        // and ar2[]        int i = 0, j = 0;         // Initialize result and current        // sum through ar1[] and ar2[].        int result = 0, sum1 = 0, sum2 = 0;         // Below 3 loops are similar to        // merge in merge sort        while (i < m && j < n) {            // Add elements of ar1[] to sum1            if (ar1[i] < ar2[j])                sum1 += ar1[i++];             // Add elements of ar2[] to sum2            else if (ar1[i] > ar2[j])                sum2 += ar2[j++];             // we reached a common point            else {                 // Take the maximum of two sums and add to                // result                // Also add the common element of array,                // once                result += max(sum1, sum2) + ar1[i];                 // Update sum1 and sum2 for elements after                // this intersection point                sum1 = 0;                sum2 = 0;                 // update i and j to move to next element of                // each array                i++;                j++;            }        }         // Add remaining elements of ar1[]        while (i < m)            sum1 += ar1[i++];         // Add remaining elements of ar2[]        while (j < n)            sum2 += ar2[j++];         // Add maximum of two sums of        // remaining elements        result += max(sum1, sum2);         return result;    }     // Driver code    public static void Main()    {        int[] ar1 = { 2, 3, 7, 10, 12, 15, 30, 34 };        int[] ar2 = { 1, 5, 7, 8, 10, 15, 16, 19 };        int m = ar1.Length;        int n = ar2.Length;         // Function call        Console.Write("Maximum sum path is :"                      + maxPathSum(ar1, ar2, m, n));    }} // This code is contributed by nitin mittal.
<?php// PHP Program to find Maximum Sum// Path in Two Arrays // This function returns the sum of// elements on maximum path// from beginning to endfunction maxPathSum($ar1, $ar2, $m, $n){         // initialize indexes for    // ar1[] and ar2[]    $i = 0;    $j = 0;     // Initialize result and    // current sum through ar1[]    // and ar2[].    $result = 0;    $sum1 = 0;    $sum2 = 0;     // Below 3 loops are similar    // to merge in merge sort    while ($i < $m and $j < $n)    {                 // Add elements of        // ar1[] to sum1        if ($ar1[$i] < $ar2[$j])            $sum1 += $ar1[$i++];         // Add elements of        // ar2[] to sum2        else if ($ar1[$i] > $ar2[$j])            $sum2 += $ar2[$j++];         // we reached a        // common point        else        {                         // Take the maximum of two sums and add to            // result              //Also add the common element of array, once            $result += max($sum1, $sum2) + $ar1[$i];             // Update sum1 and sum2 for elements after this            // intersection point            $sum1 = 0;            $sum2 = 0;                         //update i and j to move to next element of each array            $i++;            $j++;                    }    }     // Add remaining elements of ar1[]    while ($i < $m)        $sum1 += $ar1[$i++];     // Add remaining elements of ar2[]    while ($j < $n)        $sum2 += $ar2[$j++];     // Add maximum of two sums    // of remaining elements    $result += max($sum1, $sum2);     return $result;}     // Driver Code    $ar1 = array(2, 3, 7, 10, 12, 15, 30, 34);    $ar2 = array(1, 5, 7, 8, 10, 15, 16, 19);    $m = count($ar1);    $n = count($ar2);     // Function call    echo "Maximum sum path is "        , maxPathSum($ar1, $ar2, $m, $n);         // This code is contributed by anuj_67.?>
<script> // Javascript program to find maximum sum path // Utility function to find maximum of two integersfunction max(x, y){    return (x > y) ? x : y;} // This function returns the sum of elements// on maximum path from beginning to endfunction maxPathSum(ar1, ar2, m, n){         // Initialize indexes for ar1[] and ar2[]    let i = 0, j = 0;     // Initialize result and current sum    // through ar1[] and ar2[].    let result = 0, sum1 = 0, sum2 = 0;     // Below3 loops are similar to    // merge in merge sort    while (i < m && j < n)    {                 // Add elements of ar1[] to sum1        if (ar1[i] < ar2[j])            sum1 += ar1[i++];         // Add elements of ar2[] to sum2        else if (ar1[i] > ar2[j])            sum2 += ar2[j++];                     // We reached a common point        else        {            // Take the maximum of two sums and add to            // result              //Also add the common element of array, once            result += Math.max(sum1, sum2) + ar1[i];             // Update sum1 and sum2 for elements after this            // intersection point            sum1 = 0;              sum2 = 0;                         //update i and j to move to next element of each array              i++;              j++;                      }    }     // Add remaining elements of ar1[]    while (i < m)        sum1 += ar1[i++];     // Add remaining elements of ar2[]    while (j < n)        sum2 += ar2[j++];     // Add maximum of two sums of    // remaining elements    result += Math.max(sum1, sum2);     return result;} // Driver codelet ar1 = [ 2, 3, 7, 10, 12, 15, 30, 34 ];let ar2 = [ 1, 5, 7, 8, 10, 15, 16, 19 ];let m = ar1.length;let n = ar2.length; // Function calldocument.write("Maximum sum path is " +   maxPathSum(ar1, ar2, m, n));     // This code is contributed by Mayank Tyagi </script>
Maximum sum path is 122
Complexity Analysis:  
Space Complexity: O(1). Any extra space is not required, so the space complexity is constant.
Time complexity: O(m+n). In every iteration of while loops, an element from either of the two arrays is processed. There are total m + n elements. Therefore, the time complexity is O(m+n).
This article is contributed by Piyush Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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