51nod 1268 maximum distance

1272 Maximum distance 

Topic source:  Codility

Base Time Limit: 1 second Space Limit: 131072 KB Score: 20Difficulty  : Level 3 Algorithm Questions

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Given an integer array A of length N, for each array element, if there is a number greater than or equal to the element after it, these two numbers can form a pair. Each element and itself can also form a pair. For example: {5, 3, 6, 3, 4, 2}, 11 pairs can be formed as follows (numbers are subscripts):

(0,0), (0, 2), (1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3), (3 , 4), (4, 4), (5, 5). where (1, 4) is the pair with the largest distance, and the distance is 3.

Input

Line 1: 1 number N, representing the length of the array (2 <= N <= 50000).
Lines 2 - N + 1: 1 number per line, corresponding to array element Ai (1 <= Ai <= 10^9).

Output

Output the maximum distance.

Input example

6
5
3
6
3
4
2

Output example

3 


Traverse each number ai and find the farthest position on the right that is larger than ai. Sort + Suffix. Sort first, then the ith row is larger than the ith row, and then find the maximum row of the subscript in in.

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 5e4+10;
 5 int pre[N], n;
 6 struct Nod{
 7     int num,id;
 8 }nod[N];
 9 bool cmp(Nod &a, Nod &b) {
10     if(a.num != b.num) return a.num < b.num;
11     else return a.id < b.id;
12 }
13 int main() {
14     cin >> n;
15     for(int i = 1; i <= n; i ++) {
16         cin >> nod[i].num;
17         nod[i].id = i;
18     }
19     sort(nod+1,nod+1+n,cmp);
20     pre[n] = nod[n].id;
21     for(int i = n-1; i > 0; i --) {
22         pre[i] = max(pre[i+1],nod[i].id);
23     }
24     int MAX = 0;
25     for(int i = 1; i < n; i ++) {
26         MAX = max(MAX, pre[i+1]-nod[i].id);
27     }
28     cout << MAX << endl;
29     return 0;
30 }