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Maximize score of same-indexed subarrays selected from two given arrays

• Difficulty Level : Medium
• Last Updated : 03 Aug, 2021

Given two arrays A[] and B[], both consisting of N positive integers, the task is to find the maximum score among all possible same-indexed subarrays in both the arrays such that the score of any subarray over the range [L, R] is calculated by the maximum of the values (AL*BL + AL + 1*BL + 1 + … + AR*BR) + (AR*BL + AR – 1*BL + 1 + … + AL*BR).

Examples:

Input: A[] = {13, 4, 5}, B[] = {10, 22, 2}
Output: 326
Explanation:
Consider the subarrays {A[0], A[1]} and {B[0], B[1]}. Score of these subarrays can be calculated as the maximum of the following two expressions:

1. The value of the expression (A0*B0 + A1*B1) = 13 * 1 + 4 * 22 = 218.
2. The value of the expression (A0*B1 + A1*B0) = 13 * 1 + 4 * 22 = 326.

Therefore, the maximum value from the above two expressions is 326, which is the maximum score among all possible subarrays.

Input: A[] = {9, 8, 7, 6, 1}, B[]={6, 7, 8, 9, 1}
Output: 230

Naive Approach: The simplest approach to solve the given problem is to generate all possible corresponding subarray and store all the scores of all subarray generated using the given criteria. After storing all the scores, print the maximum value among all the scores generated.

Below is the implementation of the above approach:

C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to calculate the score``// of same-indexed subarrays selected``// from the arrays a[] and b[]``int` `currSubArrayScore(``int``* a, ``int``* b,``                      ``int` `l, ``int` `r)``{``    ``int` `straightScore = 0;``    ``int` `reverseScore = 0;` `    ``// Traverse the current subarray``    ``for` `(``int` `i = l; i <= r; i++) {` `        ``// Finding the score without``        ``// reversing the subarray``        ``straightScore += a[i] * b[i];` `        ``// Calculating the score of``        ``// the reversed subarray``        ``reverseScore += a[r - (i - l)]``                        ``* b[i];``    ``}` `    ``// Return the score of subarray``    ``return` `max(straightScore,``               ``reverseScore);``}` `// Function to find the subarray with``// the maximum score``void` `maxScoreSubArray(``int``* a, ``int``* b,``                      ``int` `n)``{``    ``// Stores the maximum score and the``    ``// starting and the ending point``    ``// of subarray with maximum score``    ``int` `res = 0, start = 0, end = 0;` `    ``// Traverse all the subarrays``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = i; j < n; j++) {` `            ``// Store the score of the``            ``// current subarray``            ``int` `currScore``                ``= currSubArrayScore(``                    ``a, b, i, j);` `            ``// Update the maximum score``            ``if` `(currScore > res) {``                ``res = currScore;``                ``start = i;``                ``end = j;``            ``}``        ``}``    ``}` `    ``// Print the maximum score``    ``cout << res;``}` `// Driver Code``int` `main()``{``    ``int` `A[] = { 13, 4, 5 };``    ``int` `B[] = { 10, 22, 2 };``    ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);``    ``maxScoreSubArray(A, B, N);` `    ``return` `0;``}`

Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG{` `// Function to calculate the score``// of same-indexed subarrays selected``// from the arrays a[] and b[]``static` `int` `currSubArrayScore(``int``[] a, ``int``[] b,``                             ``int` `l, ``int` `r)``{``    ``int` `straightScore = ``0``;``    ``int` `reverseScore = ``0``;` `    ``// Traverse the current subarray``    ``for``(``int` `i = l; i <= r; i++)``    ``{``        ` `        ``// Finding the score without``        ``// reversing the subarray``        ``straightScore += a[i] * b[i];` `        ``// Calculating the score of``        ``// the reversed subarray``        ``reverseScore += a[r - (i - l)] * b[i];``    ``}` `    ``// Return the score of subarray``    ``return` `Math.max(straightScore, reverseScore);``}` `// Function to find the subarray with``// the maximum score``static` `void` `maxScoreSubArray(``int``[] a, ``int``[] b, ``int` `n)``{``    ` `    ``// Stores the maximum score and the``    ``// starting and the ending point``    ``// of subarray with maximum score``    ``int` `res = ``0``, start = ``0``, end = ``0``;` `    ``// Traverse all the subarrays``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``for``(``int` `j = i; j < n; j++)``        ``{``            ` `            ``// Store the score of the``            ``// current subarray``            ``int` `currScore = currSubArrayScore(a, b, i, j);` `            ``// Update the maximum score``            ``if` `(currScore > res)``            ``{``                ``res = currScore;``                ``start = i;``                ``end = j;``            ``}``        ``}``    ``}` `    ``// Print the maximum score``    ``System.out.print(res);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `A[] = { ``13``, ``4``, ``5` `};``    ``int` `B[] = { ``10``, ``22``, ``2` `};``    ``int` `N = A.length;``    ` `    ``maxScoreSubArray(A, B, N);``}``}` `// This code is contributed by subhammahato348`

Python3

 `# Python program for the above approach` `# Function to calculate the score``# of same-indexed subarrays selected``# from the arrays a[] and b[]``def` `currSubArrayScore(a, b,``                      ``l, r):``                          ` `    ``straightScore ``=` `0``    ``reverseScore ``=` `0` `    ``# Traverse the current subarray``    ``for` `i ``in` `range``(l, r``+``1``) :` `        ``# Finding the score without``        ``# reversing the subarray``        ``straightScore ``+``=` `a[i] ``*` `b[i]` `        ``# Calculating the score of``        ``# the reversed subarray``        ``reverseScore ``+``=` `a[r ``-` `(i ``-` `l)] ``*` `b[i]``    `  `    ``# Return the score of subarray``    ``return` `max``(straightScore,``               ``reverseScore)`  `# Function to find the subarray with``# the maximum score``def` `maxScoreSubArray(a, b,``                      ``n) :``                          ` `    ``# Stores the maximum score and the``    ``# starting and the ending point``    ``# of subarray with maximum score``    ``res ``=` `0``    ``start ``=` `0``    ``end ``=` `0` `    ``# Traverse all the subarrays``    ``for` `i ``in` `range``(n) :``        ``for` `j ``in` `range``(i, n) :` `            ``# Store the score of the``            ``# current subarray``            ``currScore ``=` `currSubArrayScore(a, b, i, j)` `            ``# Update the maximum score``            ``if` `(currScore > res) :``                ``res ``=` `currScore``                ``start ``=` `i``                ``end ``=` `j` `    ``# Prthe maximum score``    ``print``(res)` `# Driver Code` `A ``=` `[ ``13``, ``4``, ``5` `]``B ``=` `[ ``10``, ``22``, ``2` `]``N ``=` `len``(A)``maxScoreSubArray(A, B, N)` `# This code is contributed by target_2.`

C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to calculate the score``// of same-indexed subarrays selected``// from the arrays a[] and b[]``static` `int` `currSubArrayScore(``int``[] a, ``int``[] b,``                             ``int` `l, ``int` `r)``{``    ``int` `straightScore = 0;``    ``int` `reverseScore = 0;`` ` `    ``// Traverse the current subarray``    ``for``(``int` `i = l; i <= r; i++)``    ``{``        ` `        ``// Finding the score without``        ``// reversing the subarray``        ``straightScore += a[i] * b[i];`` ` `        ``// Calculating the score of``        ``// the reversed subarray``        ``reverseScore += a[r - (i - l)] * b[i];``    ``}`` ` `    ``// Return the score of subarray``    ``return` `Math.Max(straightScore, reverseScore);``}`` ` `// Function to find the subarray with``// the maximum score``static` `void` `maxScoreSubArray(``int``[] a, ``int``[] b, ``int` `n)``{``    ` `    ``// Stores the maximum score and the``    ``// starting and the ending point``    ``// of subarray with maximum score``    ``int` `res = 0;`` ` `    ``// Traverse all the subarrays``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``for``(``int` `j = i; j < n; j++)``        ``{``            ` `            ``// Store the score of the``            ``// current subarray``            ``int` `currScore = currSubArrayScore(``                ``a, b, i, j);`` ` `            ``// Update the maximum score``            ``if` `(currScore > res)``            ``{``                ``res = currScore;``            ``}``        ``}``    ``}``    ` `    ``// Print the maximum score``    ``Console.Write(res);``}`` ` `// Driver Code``static` `public` `void` `Main()``{``    ``int``[] A = { 13, 4, 5 };``    ``int``[] B = { 10, 22, 2 };``    ``int` `N = A.Length;``    ` `    ``maxScoreSubArray(A, B, N);``}``}` `// This code is contributed by unknown2108`

Javascript

 ``
Output:
`326`

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by considering every element as the middle point of every possible subarray and then expand the subarray in both directions while updating the maximum score for every value. Follow the steps below to solve the problem:

• Initialize a variable, say res to store the resultant maximum value.
• Iterate over the range [1, N – 1] using the variable mid and perform the following steps:
• Initialize two variables, say score1 and score2 as A[mid]*B[mid] in case of odd length subarray.
• Initialize two variables, say prev as (mid – 1) and next as (mid + 1) to expand the current subarray.
• Iterate a loop until prev is positive and the value of next is less than N and perform the following steps:
• Add the value of (a[prev]*b[prev]+a[next]*b[next]) to the variable score1.
• Add the value of (a[prev]*b[next]+a[next]*b[prev]) to the variable score2.
• Update the value of res to the maximum of score1, score2, and res.
• Decrement the value of prev by 1 and increment the value of next by 1.
• Update the value of score1 and score2 as 0 and set the value of prev as (mid – 1) and next as mid to consider the case of even length subarray.
• After completing the above steps, print the value of res as the result.

Below is the implementation of the above approach:

C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to calculate the score``// of same-indexed subarrays selected``// from the arrays a[] and b[]``void` `maxScoreSubArray(``int``* a, ``int``* b,``                      ``int` `n)``{``    ``// Store the required result``    ``int` `res = 0;` `    ``// Iterate in the range [0, N-1]``    ``for` `(``int` `mid = 0; mid < n; mid++) {` `        ``// Consider the case of odd``        ``// length subarray``        ``int` `straightScore = a[mid] * b[mid],``            ``reverseScore = a[mid] * a[mid];``        ``int` `prev = mid - 1, next = mid + 1;` `        ``// Update the maximum score``        ``res = max(res, max(straightScore,``                           ``reverseScore));` `        ``// Expanding the subarray in both``        ``// directions with equal length``        ``// so that mid point remains same``        ``while` `(prev >= 0 && next < n) {` `            ``// Update both the scores``            ``straightScore``                ``+= (a[prev] * b[prev]``                    ``+ a[next] * b[next]);``            ``reverseScore``                ``+= (a[prev] * b[next]``                    ``+ a[next] * b[prev]);` `            ``res = max(res,``                      ``max(straightScore,``                          ``reverseScore));` `            ``prev--;``            ``next++;``        ``}` `        ``// Consider the case of``        ``// even length subarray``        ``straightScore = 0;``        ``reverseScore = 0;` `        ``prev = mid - 1, next = mid;``        ``while` `(prev >= 0 && next < n) {` `            ``// Update both the scores``            ``straightScore``                ``+= (a[prev] * b[prev]``                    ``+ a[next] * b[next]);``            ``reverseScore``                ``+= (a[prev] * b[next]``                    ``+ a[next] * b[prev]);` `            ``res = max(res,``                      ``max(straightScore,``                          ``reverseScore));` `            ``prev--;``            ``next++;``        ``}``    ``}` `    ``// Print the result``    ``cout << res;``}` `// Driver Code``int` `main()``{``    ``int` `A[] = { 13, 4, 5 };``    ``int` `B[] = { 10, 22, 2 };``    ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);``    ``maxScoreSubArray(A, B, N);` `    ``return` `0;``}`

Java

 `// Java Program for the above approach``import` `java.io.*;` `class` `GFG {``    ``// Function to calculate the score``    ``// of same-indexed subarrays selected``    ``// from the arrays a[] and b[]``    ``static` `void` `maxScoreSubArray(``int``[] a, ``int``[] b, ``int` `n)``    ``{``        ``// Store the required result``        ``int` `res = ``0``;` `        ``// Iterate in the range [0, N-1]``        ``for` `(``int` `mid = ``0``; mid < n; mid++) {` `            ``// Consider the case of odd``            ``// length subarray``            ``int` `straightScore = a[mid] * b[mid],``                ``reverseScore = a[mid] * a[mid];``            ``int` `prev = mid - ``1``, next = mid + ``1``;` `            ``// Update the maximum score``            ``res = Math.max(``                ``res, Math.max(straightScore, reverseScore));` `            ``// Expanding the subarray in both``            ``// directions with equal length``            ``// so that mid point remains same``            ``while` `(prev >= ``0` `&& next < n) {` `                ``// Update both the scores``                ``straightScore += (a[prev] * b[prev]``                                  ``+ a[next] * b[next]);``                ``reverseScore += (a[prev] * b[next]``                                 ``+ a[next] * b[prev]);` `                ``res = Math.max(res, Math.max(straightScore,``                                             ``reverseScore));` `                ``prev--;``                ``next++;``            ``}` `            ``// Consider the case of``            ``// even length subarray``            ``straightScore = ``0``;``            ``reverseScore = ``0``;` `            ``prev = mid - ``1``;``            ``next = mid;``            ``while` `(prev >= ``0` `&& next < n) {` `                ``// Update both the scores``                ``straightScore += (a[prev] * b[prev]``                                  ``+ a[next] * b[next]);``                ``reverseScore += (a[prev] * b[next]``                                 ``+ a[next] * b[prev]);` `                ``res = Math.max(res, Math.max(straightScore,``                                             ``reverseScore));` `                ``prev--;``                ``next++;``            ``}``        ``}` `        ``// Print the result``        ``System.out.println(res);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `A[] = { ``13``, ``4``, ``5` `};``        ``int` `B[] = { ``10``, ``22``, ``2` `};``        ``int` `N = A.length;``        ``maxScoreSubArray(A, B, N);``    ``}``}` `        ``// This code is contributed by Potta Lokesh`

Python3

 `# Python3 program for the above approach` `# Function to calculate the score``# of same-indexed subarrays selected``# from the arrays a[] and b[]``def` `maxScoreSubArray(a, b, n):``    ` `    ``# Store the required result``    ``res ``=` `0` `    ``# Iterate in the range [0, N-1]``    ``for` `mid ``in` `range``(n):``        ` `        ``# Consider the case of odd``        ``# length subarray``        ``straightScore ``=` `a[mid] ``*` `b[mid]``        ``reverseScore ``=` `a[mid] ``*` `a[mid]``        ``prev ``=` `mid ``-` `1``        ``next` `=` `mid ``+` `1` `        ``# Update the maximum score``        ``res ``=` `max``(res, ``max``(straightScore,``                           ``reverseScore))` `        ``# Expanding the subarray in both``        ``# directions with equal length``        ``# so that mid poremains same``        ``while` `(prev >``=` `0` `and` `next` `< n):` `            ``# Update both the scores``            ``straightScore ``+``=` `(a[prev] ``*` `b[prev] ``+``                              ``a[``next``] ``*` `b[``next``])``            ``reverseScore ``+``=` `(a[prev] ``*` `b[``next``] ``+``                             ``a[``next``] ``*` `b[prev])` `            ``res ``=` `max``(res, ``max``(straightScore,``                               ``reverseScore))` `            ``prev ``-``=` `1``            ``next` `+``=` `1` `        ``# Consider the case of``        ``# even length subarray``        ``straightScore ``=` `0``        ``reverseScore ``=` `0` `        ``prev ``=` `mid ``-` `1``        ``next` `=` `mid``        ` `        ``while` `(prev >``=` `0` `and` `next` `< n):` `            ``# Update both the scores``            ``straightScore ``+``=` `(a[prev] ``*` `b[prev] ``+``                              ``a[``next``] ``*` `b[``next``])``            ``reverseScore ``+``=` `(a[prev] ``*` `b[``next``] ``+``                             ``a[``next``] ``*` `b[prev])` `            ``res ``=` `max``(res, ``max``(straightScore,``                               ``reverseScore))` `            ``prev ``-``=` `1``            ``next` `+``=` `1` `    ``# Print the result``    ``print``(res)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``A ``=` `[ ``13``, ``4``, ``5` `]``    ``B ``=` `[ ``10``, ``22``, ``2` `]``    ``N ``=` `len``(A)``    ` `    ``maxScoreSubArray(A, B, N)` `# This code is contributed by mohit kumar 29`

C#

 `// C# Program for the above approach``using` `System;` `public` `class` `GFG{``    ` `    ``// Function to calculate the score``    ``// of same-indexed subarrays selected``    ``// from the arrays a[] and b[]``    ``static` `void` `maxScoreSubArray(``int``[] a, ``int``[] b, ``int` `n)``    ``{``      ` `        ``// Store the required result``        ``int` `res = 0;`` ` `        ``// Iterate in the range [0, N-1]``        ``for` `(``int` `mid = 0; mid < n; mid++) {`` ` `            ``// Consider the case of odd``            ``// length subarray``            ``int` `straightScore = a[mid] * b[mid],``                ``reverseScore = a[mid] * a[mid];``            ``int` `prev = mid - 1, next = mid + 1;`` ` `            ``// Update the maximum score``            ``res = Math.Max(``                ``res, Math.Max(straightScore, reverseScore));`` ` `            ``// Expanding the subarray in both``            ``// directions with equal length``            ``// so that mid point remains same``            ``while` `(prev >= 0 && next < n) {`` ` `                ``// Update both the scores``                ``straightScore += (a[prev] * b[prev]``                                  ``+ a[next] * b[next]);``                ``reverseScore += (a[prev] * b[next]``                                 ``+ a[next] * b[prev]);`` ` `                ``res = Math.Max(res, Math.Max(straightScore,``                                             ``reverseScore));`` ` `                ``prev--;``                ``next++;``            ``}`` ` `            ``// Consider the case of``            ``// even length subarray``            ``straightScore = 0;``            ``reverseScore = 0;`` ` `            ``prev = mid - 1;``            ``next = mid;``            ``while` `(prev >= 0 && next < n) {`` ` `                ``// Update both the scores``                ``straightScore += (a[prev] * b[prev]``                                  ``+ a[next] * b[next]);``                ``reverseScore += (a[prev] * b[next]``                                 ``+ a[next] * b[prev]);`` ` `                ``res = Math.Max(res, Math.Max(straightScore,``                                             ``reverseScore));`` ` `                ``prev--;``                ``next++;``            ``}``        ``}`` ` `        ``// Print the result``        ``Console.WriteLine(res);``    ``}`` ` `    ``// Driver Code  ``    ``static` `public` `void` `Main (){``        ` `        ``int``[] A = { 13, 4, 5 };``        ``int``[] B = { 10, 22, 2 };``        ``int` `N = A.Length;``        ``maxScoreSubArray(A, B, N);``        ` `    ``}``}` `// This code is contributed by patel2127.`

Javascript

 ``
Output:
`326`

Time Complexity: O(N2)
Auxiliary Space: O(1)

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