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Trapping Rain Water.cpp
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Trapping Rain Water.cpp
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/*
Solution by Rahul Surana
***********************************************************
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.
Example 1:
| ___
| ___ | |___ ___
| | |~~~| |~~~| |
------------------------------------
|
Input: height = [0,1,0,2,1,0,1]
Output: 2
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1]. In this case, 2 units of rain water (blue section) are being trapped.
***********************************************************
*/
#include <bits/stdc++.h>
#define ll long long
#define vl vector<ll>
#define vi vector<int>
#define pi pair<int,int>
#define pl pair<ll,ll>
#define all(a) a.begin(),a.end()
#define mem(a,x) memset(a,x,sizeof(a))
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define FOR(i,a) for(int i = 0; i < a; i++)
#define trace(x) cerr<<#x<<" : "<<x<<endl;
#define trace2(x,y) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<endl;
#define trace3(x,y,z) cerr<<#x<<" : "<<x<<" | "<<#y<<" : "<<y<<" | "<<#z<<" : "<<z<<endl;
#define fast_io std::ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
#define MOD 1000000007
using namespace std;
int trap(vector<int>& height) {
int n = height.size();
if(n<=2) return 0;
vector<int> prefix(n,0), suffix(n,0);
prefix[0] = height[0];
for(int i = 1; i < n; i++){
prefix[i] = max(height[i],prefix[i-1]);
}
suffix[n-1] = height[n-1];
for(int i = n-2; i >=0 ; i--){
suffix[i] = max(height[i],suffix[i+1]);
}
int ans = 0;
for(int i = 1; i < n-1; i++){
ans+= min(prefix[i],suffix[i]) - height[i];
}
return ans;
}
int main()
{
fast_io;
int n;
vector<int> ar(n,0);
FOR(i,n) cin >> ar[i];
cout << trap(ar)<<"\n";
}