# Maximum Sum Path in Two Arrays

Given two sorted arrays such the arrays may have some common elements. Find the sum of the maximum sum path to reach from 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 that the common elements do not have to be at same indexes.

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

Input:  ar1[] = {10, 12}, ar2 = {5, 7, 9}
Output: 22
22 is 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
122 is sum of 1, 5, 7, 8, 10, 12, 15, 30, 34
```

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

The idea is to do something similar to merge process of merge sort. We need to calculate sums of elements between all common points for both arrays. Whenever we see a common point, we compare the two sums and add the maximum of two to the result. Following are detailed steps.

1) 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.

2) Now run a loop to traverse elements of both arrays. While traversing compare current elements of ar1[] and ar2[].

2.a) If current element of ar1[] is smaller than current element of ar2[], then update sum1, else if current element of ar2[] is smaller, then update sum2.

2.b) If current element of ar1[] and ar2[] are same, then take the maximum of sum1 and sum2 and add it to the result. Also add the common element to the result.

Following is the implementation of above approach.

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// 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++]; ` ` `  `        ``else`  `// we reached a common point ` `        ``{ ` `            ``// Take the maximum of two sums and add to result ` `            ``result += max(sum1, sum2); ` ` `  `            ``// Update sum1 and sum2 for elements after this ` `            ``// intersection point ` `            ``sum1 = 0, sum2 = 0; ` ` `  `            ``// Keep updating result while there are more common ` `            ``// elements ` `            ``while` `(i < m &&  j < n && ar1[i] == ar2[j]) ` `            ``{ ` `                ``result = result + ar1[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 program to test above function ` `int` `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); ` `    ``int` `n = ``sizeof``(ar2)/``sizeof``(ar2); ` `    ``cout << ``"Maximum sum path is "`  `         ``<< maxPathSum(ar1, ar2, m, n); ` `    ``return` `0; ` `} `

## Java

 `class` `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 ` `                ``result += max(sum1, sum2); ` ` `  `                ``// Update sum1 and sum2 for elements after this ` `                ``// intersection point ` `                ``sum1 = ``0``; ` `                ``sum2 = ``0``; ` ` `  `                ``// Keep updating result while there are more common ` `                ``// elements ` `                ``while` `(i < m && j < n && ar1[i] == ar2[j])  ` `                ``{ ` `                    ``result = result + ar1[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 program to test above functions ` `    ``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; ` `        ``System.out.println(``"Maximum sum path is :"` `+  ` `                                       ``sumpath.maxPathSum(ar1, ar2, m, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by Mayank Jaiswal `

## Python

 `# 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 sum1     ` `        ``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) ` `             `  `            ``# Update sum1 and sum2 for elements after this intersection point ` `            ``sum1, sum2 ``=` `0``, ``0` `             `  `            ``# Keep updating result while there are more common elements ` `            ``while` `(i < m ``and` `j < n ``and` `ar1[i]``=``=``ar2[j]): ` `                ``result ``+``=` `ar1[i] ` `                ``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 function ` `ar1 ``=` `[``2``, ``3``, ``7``, ``10``, ``12``, ``15``, ``30``, ``34``] ` `ar2 ``=` `[``1``, ``5``, ``7``, ``8``, ``10``, ``15``, ``16``, ``19``] ` `m ``=` `len``(ar1) ` `n ``=` `len``(ar2) ` `print` `"Maximum sum path is"``, maxPathSum(ar1, ar2, m, n) ` ` `  `# This code is contributed by __Devesh Agrawal__ `

## C#

 `// C# program for Maximum Sum Path in ` `// Two Arrays ` `using` `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 ` `                ``result += max(sum1, sum2); ` ` `  `                ``// Update sum1 and sum2 for ` `                ``// elements after this ` `                ``// intersection point ` `                ``sum1 = 0; ` `                ``sum2 = 0; ` ` `  `                ``// Keep updating result while  ` `                ``// there are more common ` `                ``// elements ` `                ``while` `(i < m && j < n &&  ` `                            ``ar1[i] == ar2[j])  ` `                ``{ ` `                    ``result = result + ar1[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 program to test above functions ` `    ``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; ` `        ``Console.Write(``"Maximum sum path is :"` `+  ` `                     ``maxPathSum(ar1, ar2, m, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` ``\$ar2``[``\$j``]) ` `            ``\$sum2` `+= ``\$ar2``[``\$j``++]; ` ` `  `        ``// we reached a  ` `        ``// common point ` `        ``else`  `        ``{ ` `             `  `            ``// Take the maximum of two ` `            ``// sums and add to result ` `            ``\$result` `+= max(``\$sum1``, ``\$sum2``); ` ` `  `            ``// Update sum1 and sum2 for ` `            ``// elements after this ` `            ``// intersection point ` `            ``\$sum1` `= 0;  ` `            ``\$sum2` `= 0; ` ` `  `            ``// Keep updating result while ` `            ``// there are more common ` `            ``// elements ` `            ``while` `(``\$i` `< ``\$m` `and` `\$j` `< ``\$n` `and`  `                   ``\$ar1``[``\$i``] == ``\$ar2``[``\$j``]) ` `            ``{ ` `                ``\$result` `= ``\$result` `+ ``\$ar1``[``\$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``); ` `    ``echo` `"Maximum sum path is "` `        ``, maxPathSum(``\$ar1``, ``\$ar2``, ``\$m``, ``\$n``); ` `         `  `// This code is contributed by anuj_67. ` `?> `

Output:

`Maximum sum path is 122`

Time complexity: In every iteration of while loops, we process an element from either of the two arrays. There are total m + n elements. Therefore, time complexity is O(m+n).

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Improved By : nitin mittal, vt_m

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