please give me a hint to solve these question.Asap!
I didnt get the question well
EXPLANATION
An element of the array can store water if there are higher bars on left and right. We can find the amount of water to be stored in every element by finding the heights of bars on left and right sides. The idea is to compute the amount of water that can be stored in every element of array. For example, consider the array {3, 0, 0, 2, 0, 4}, we can store three units of water at indexes 1 and 2, and one unit of water at index 3, and three units of water at index 4.
A Simple Solution is to traverse every array element and find the highest bars on left and right sides. Take the smaller of two heights. The difference between the smaller height and height of the current element is the amount of water that can be stored in this array element. Time complexity of this solution is O(n2).
Naive Approach
int maxWater(int arr[], int n)
{
// To store the maximum water
// that can be stored
int res = 0;
// For every element of the array
for (int i = 1; i < n-1; i++) {
// Find the maximum element on its left
int left = arr[i];
for (int j=0; j<i; j++)
left = max(left, arr[j]);
// Find the maximum element on its right
int right = arr[i];
for (int j=i+1; j<n; j++)
right = max(right, arr[j]);
// Update the maximum water
res = res + (min(left, right) - arr[i]);
}
return res;
}
Better Approach
An element of an array can store water if there are higher bars on left and right. We can find the amount of water to be stored in every element by finding the heights of bars on the left and right sides. The idea is to compute the amount of water that can be stored in every element of the array. For example, consider the array {3, 0, 0, 2, 0, 4}, we can store two units of water at indexes 1 and 2, and one unit of water at index 2.
Pre-compute highest bar on left and right of every bar in O(n) time. Then use these pre-computed values to find the amount of water in every array element.
C++ Code
#include <iostream>
using namespace std;
int a[1000000], l[1000000], r[1000000];
int main()
{
int n, i, j;
cin >> n;
for (i = 0; i < n; i++)
{
cin >> a[i];
}
l[0] = a[0];
r[n - 1] = a[n - 1];
int leftmax = 0, rightmax = 0;
for (i = 1; i < n; i++)
{
l[i] = max(l[i - 1], a[i]);
}
for (i = n - 2; i >= 0; i--)
{
r[i] = max(r[i + 1], a[i]);
}
int water = 0;
for (i = 0; i < n; i++)
{
water += (min(l[i], r[i]) - a[i]);
}
cout << water;
return 0;
}
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