Best Sightseeing Pair LT1014

Posted 2021-02-14 taste-it-own-it-love-it

Given an array A of positive integers, A[i] represents the value of the i-th sightseeing spot, and two sightseeing spots i and j have distance j - i between them.

The score of a pair (i < j) of sightseeing spots is (A[i] + A[j] + i - j) : the sum of the values of the sightseeing spots, minus the distance between them.

Return the maximum score of a pair of sightseeing spots.

Example 1:

Input: [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11

 

Note:

1. 2 <= A.length <= 50000
2. 1 <= A[i] <= 1000

 Idea 1.  max(A[i] + A[j] + i -j) (i < j) = max(max(A[i] + i) + A[j] -j), scan from left to right

Time complexity: O(n)

Space complexity: O(1)

 1 class Solution {
 2     public int maxScoreSightseeingPair(int[] A) {
 3         int maxScore = Integer.MIN_VALUE;
 4         int sumVal = Integer.MIN_VALUE;
 5         for(int i = 1; i < A.length; ++i) {
 6             sumVal = Math.max(sumVal, A[i-1] + (i-1));
 7             maxScore = Math.max(maxScore, A[i] - i + sumVal);
 8         }
 9         
10         return maxScore;
11     }
12 }

Idea 1.b scan from right to left, max(A[i] + A[j] + i -j) (i < j)  = max(max(A[j] -j) + A[i] + i)

 1 class Solution {
 2     public int maxScoreSightseeingPair(int[] A) {
 3         int maxScore = Integer.MIN_VALUE;
 4         int sumVal = Integer.MIN_VALUE;
 5         for(int i = A.length-2; i >= 0; --i) {
 6             sumVal = Math.max(sumVal, A[i+1] - (i+1));
 7             maxScore = Math.max(maxScore, A[i] + i + sumVal);
 8         }
 9         
10         return maxScore;
11     }
12 }