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.
Implementation:
C++
#include<iostream> 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[0]); int n = sizeof(ar2)/sizeof(ar2[0]); cout << "Maximum sum path is " << maxPathSum(ar1, ar2, m, n); return 0; } |
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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 |
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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__ |
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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. |
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PHP
<?php // PHP Program to find Maximum Sum // Path in Two Arrays // This function returns the sum of // elements on maximum path // from beginning to end function 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 $result += max($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. ?> |
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Output:
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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