Given a non-negative integer array representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining. For example, given [0,1,0,2,1,0,1,3,2,1,2,1], returns 6.

Follow up:

What is the complexity of your solution?

for input:

A=[0,1,0,2,1,0,1,3,2,1,2,1]

N= sizeof(A)

compute Left[] where Left[i]= Max(A[0..i]) Calculate Max bar height from node 0-i

Left=[0,1,1,2,2,2,2,3,3,3,3,3]

compute Right[] where Right[i]= Max(A[i..N]) Calculate Max bar height from node i-n

Right=[3,3,3,3,3,3,3,3,2,2,2,1]

compute D[] where D[i]=Min(Left[i],Right[i]) Calculate Min bar height between bar before Vs after i, coz only this much water could fill

D=[0,1,1,2,2,2,2,3,2,2,2,1]

compute E[] where E[i]=D[i]-A[i] Subtract the height of bar i from Max water bars could hold before Vs after

E[i]=[0,0,1,0,1,2,1,0,0,1,0,0]

the answer is sum(E[i])

the complexity is O(N)

public class Solution {

public int trap(int[] A) {

// Start typing your Java solution below

// DO NOT write main() function

int[]left=new int[A.length];

int[]right=new int[A.length];

if(A.length==0) return 0;

int max=0;

for(int i =0;i<A.length;i++){

max=Math.max(max,A[i]);

left[i]=max;

}

max=0;

for(int i =A.length-1;i>=0;i--){

max=Math.max(max,A[i]);

right[i]=max;

}

max=0;

for(int i =0;i<A.length;i++)

max+=Math.min(left[i],right[i])-A[i];

return max;

}

}

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