Find the Best Sightseeing Pair

Last Updated : 31 May, 2024

Given an integer array arr[] where arr[i] represents the value of the ith sightseeing spot. Two sightseeing spots i and j have a distance j - i between them. The score of a pair (i < j) of sightseeing spots is arr[i] + arr[j] + i - j: the sum of arr[]of the sightseeing spots, minus the distance between them. Find the maximum score of a pair of sightseeing spots.

Examples:

Input: arr[]= {8, 1, 5, 2, 6}
Output: 11
Explanation: For i = 0, j = 2, arr[i] + arr[j] + i - j = 8 + 5 + 0 - 2 = 11

Input: arr= [1, 2]
Output: 2
Explanation: For i = 0, j = 1, arr[i] + arr[j] + i - j = 2 + 1 + 0 - 1 = 2

Approach: To solve the problem, follow the below idea:

Since our task is to find max{ (arr[i]+i) + (arr[j]-j) } for all i<j.

So we can say that we want maximum of arr[i]+i and maximum of arr[j]-j.

Subproblem: find max (arr[i]-i) after i.
Recurrence Relation: max_after(i) = max{ arr[i]-i, max_after(i+1) }.
Finally ans would be maximum of arr[i]+i+max_after(i+1).

Step-by-step algorithm:

Maintain an array max_after[], such that max_after[i] stores the maximum value of arr[j] - j among all j >= i.
Now, iterate from 0 to N - 1, and for every index i store the maximum value of arr[i] + i + max_after[i] and store it in ans.
Return ans as the final answer.

Below is the implementation of the above algorithm:

C++

#include <bits/stdc++.h> using namespace std;  int maxScoreSightseeingPair(vector<int>& arr) {     // This function finds the best sightseeing pair in a     // city represented by an array values.     //  - max_after: stores the maximum sightseeing score     //  obtainable starting from city i+1      int N = arr.size();     vector<int> max_after(N, 0);      // Initialize max_after for the last city (no city after     // it).     max_after[N - 1] = arr[N - 1] - (N - 1);      // Fill max_after array in reverse order.     for (int i = N - 2; i >= 0; i--) {         max_after[i] = max(max_after[i + 1], arr[i] - i);     }      int ans = 0;     // Iterate through all cities except the last one.     for (int i = 0; i < N - 1; i++) {         // Calculate the total sightseeing score for the         // current city and its best pairing city (i+1).         ans = max(ans, arr[i] + i + max_after[i + 1]);     }      return ans; }  int main() {     // Driver code to test the function     vector<int> arr = { 8, 1, 5, 2, 6 };      int maxScore = maxScoreSightseeingPair(arr);     cout << "Max sightseeing score: " << maxScore << endl;     return 0; }

Java

import java.util.Arrays;  public class MaxScoreSightseeingPair {     public static int maxScoreSightseeingPair(int[] arr)     {         // This function finds the best sightseeing pair in         // a city represented by an array values.         //  - maxAfter: stores the maximum sightseeing score         //  obtainable starting from city i+1          int N = arr.length;         int[] maxAfter = new int[N];          // Initialize maxAfter for the last city (no city         // after it).         maxAfter[N - 1] = arr[N - 1] - (N - 1);          // Fill maxAfter array in reverse order.         for (int i = N - 2; i >= 0; i--) {             maxAfter[i]                 = Math.max(maxAfter[i + 1], arr[i] - i);         }          int ans = 0;         // Iterate through all cities except the last one.         for (int i = 0; i < N - 1; i++) {             // Calculate the total sightseeing score for the             // current city and its best pairing city (i+1).             ans = Math.max(ans,                            arr[i] + i + maxAfter[i + 1]);         }          return ans;     }      public static void main(String[] args)     {         // Driver code to test the function         int[] arr = { 8, 1, 5, 2, 6 };          int maxScore = maxScoreSightseeingPair(arr);         System.out.println("Max sightseeing score: "                            + maxScore);     } }  // This code is contributed by Shivam

Python

def max_score_sightseeing_pair(arr):     N = len(arr)     max_after = [0] * N      max_after[N - 1] = arr[N - 1] - (N - 1)      for i in range(N - 2, -1, -1):         max_after[i] = max(max_after[i + 1], arr[i] - i)      ans = 0     for i in range(N - 1):         ans = max(ans, arr[i] + i + max_after[i + 1])      return ans   # Driver code to test the function arr = [8, 1, 5, 2, 6] max_score = max_score_sightseeing_pair(arr) print("Max sightseeing score:", max_score)

JavaScript

// Function to calculate the maximum score for a sightseeing pair function maxScoreSightseeingPair(arr) {     // Get the length of the array     const N = arr.length;     // Initialize an array to store the maximum values after each index     const maxAfter = new Array(N).fill(0);      // Calculate the maximum value after each index     maxAfter[N - 1] = arr[N - 1] - (N - 1);     for (let i = N - 2; i >= 0; i--) {         maxAfter[i] = Math.max(maxAfter[i + 1], arr[i] - i);     }      // Initialize a variable to store the maximum score     let ans = 0;     // Iterate through the array to find the maximum score     for (let i = 0; i < N - 1; i++) {         ans = Math.max(ans, arr[i] + i + maxAfter[i + 1]);     }      // Return the maximum score     return ans; }  // Driver code to test the function const arr = [8, 1, 5, 2, 6]; const maxScore = maxScoreSightseeingPair(arr); console.log("Max sightseeing score:", maxScore);  // This code is contributed by Shivam Gupta

Output

Max sightseeing score: 11

Time complexity: O(N), where N is the size of input subarray arr[].
Auxiliary Space: O(N)

Find the Best Sightseeing Pair

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