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:
- The value of the expression (A0*B0 + A1*B1) = 13 * 1 + 4 * 22 = 218.
- 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 <bits/stdc++.h>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 scorevoid 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 Codeint 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 approachimport 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 scorestatic 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 Codepublic 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 scoredef 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 # Print the maximum score print(res)# Driver CodeA = [ 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 approachusing 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 scorestatic 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 Codestatic 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
<script>// JavaScript program for the above approach// Function to calculate the score// of same-indexed subarrays selected// from the arrays a[] and b[]function currSubArrayScore(a, b, l, r) { let straightScore = 0; let reverseScore = 0; // Traverse the current subarray for (let 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 scorefunction maxScoreSubArray(a, b, n) { // Stores the maximum score and the // starting and the ending point // of subarray with maximum score let res = 0, start = 0, end = 0; // Traverse all the subarrays for (let i = 0; i < n; i++) { for (let j = i; j < n; j++) { // Store the score of the // current subarray let currScore = currSubArrayScore( a, b, i, j); // Update the maximum score if (currScore > res) { res = currScore; start = i; end = j; } } } // Print the maximum score document.write(res);}// Driver Codelet A = [13, 4, 5];let B = [10, 22, 2];let N = A.length;maxScoreSubArray(A, B, N);// This code is contributed by gfgking.</script> |
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 <bits/stdc++.h>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 Codeint 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 approachimport 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 Codeif __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 approachusing 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
<script>// JavaScript program for the above approach// Function to calculate the score// of same-indexed subarrays selected// from the arrays a[] and b[]function maxScoreSubArray(a, b, n) { // Store the required result let res = 0; // Iterate in the range [0, N-1] for (let mid = 0; mid < n; mid++) { // Consider the case of odd // length subarray let straightScore = a[mid] * b[mid], reverseScore = a[mid] * a[mid]; let 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 document.write(res);}// Driver Codelet A = [13, 4, 5];let B = [10, 22, 2];let N = A.lengthmaxScoreSubArray(A, B, N);</script> |
Output:
326
Time Complexity: O(N2)
Auxiliary Space: O(1)
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