92 lines
2.1 KiB
Java
92 lines
2.1 KiB
Java
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// Given an array arr[] of N non-negative integers representing the height of blocks.
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// If width of each block is 1, compute how much water can be trapped between the blocks during the rainy season.
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//Time Complexity: O(N)
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//Auxiliary Space: O(1)
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import java.io.*;
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import java.util.*;
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import java.lang.*;
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class Main {
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public static void main (String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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int t = Integer.parseInt(br.readLine().trim()); //Inputting the testcases
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while(t-->0){
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//size of array
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int n = Integer.parseInt(br.readLine().trim());
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int arr[] = new int[n];
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String inputLine[] = br.readLine().trim().split(" ");
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//adding elements to the array
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for(int i=0; i<n; i++){
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arr[i] = Integer.parseInt(inputLine[i]);
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}
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Solution obj = new Solution();
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//calling trappingWater() function
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System.out.println(obj.trappingWater(arr, n));
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}
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}
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}
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class Solution{
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// arr: input array
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// n: size of array
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// Function to find the trapped water between the blocks.
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static int trappingWater(int arr[], int n) {
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// initialize output
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int result = 0;
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// maximum element on left and right
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int left_max = 0, right_max = 0;
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// indices to traverse the array
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int lo = 0, hi = n - 1;
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while (lo <= hi) {
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if (arr[lo] < arr[hi]) {
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if (arr[lo] > left_max)
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// update max in left
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left_max = arr[lo];
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else
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// water on curr element = max - curr
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result += left_max - arr[lo];
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lo++;
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}
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else {
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if (arr[hi] > right_max)
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// update right maximum
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right_max = arr[hi];
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else
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result += right_max - arr[hi];
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hi--;
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}
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}
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return result;
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}
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}
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/* Test Case:
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Input:
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N = 6
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arr[] = {3,0,0,2,0,4}
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Output:
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10
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Explanation:
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____|
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300204
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Total Water Trapped: 3+3+1+3 = 10
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*/
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