Merge pull request #482 from sumitvajarinkar/trapping-rain-water

closes #443 Trapping rain water

Author: smv1999

Arrays/trapping_rain_water_brute_force.cpp

@@ -0,0 +1,50 @@
+// Trapping Rain Water
+//Approach 1
+// O(n^2) O(1)
+
+//i/p: 3 0 1 2 5
+//o/p: 6
+
+//Brute force approach
+
+#include<bits/stdc++.h>
+using namespace std;
+int traprain(int arr[],int n)
+{ // To store the maximum water
+    // that can be stored
+    int res = 0;
+
+// i/p: 3 0 1 2 5
+//o/p: 6
+
+    // For every element of the array
+    for (int i = 1; i < n-1; i++) {
+
+        // Find the maximum element on its left from o th to i th value
+        int left = arr[i];
+        for (int j=0; j<i; j++)                 //o(n^2)
+           left = max(left, arr[j]);
+
+        // Find the maximum element on its right from j th th value to end
+        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]); // 3 + 2 + 1
+    }
+
+    return res;
+}
+int main()
+{
+    int n;
+    cout<<"Enter the number of elements in the array\n";
+    cin>>n;
+    int arr[n];
+    cout<<"Enter the elements in  the array\n";
+    for(int i=0;i<n;i++)
+        cin>>arr[i];
+    cout<<"Total trapped water is "<<traprain(arr,n);
+    return 0;
+}

Arrays/trapping_rain_water_extra_space.cpp

@@ -0,0 +1,61 @@
+// Trapping Rain Water
+//Approach 2
+// O(n) O(2n)
+
+//i/p: 3 0 1 2 5
+//o/p: 6
+
+//Use 2 extra arrays
+
+#include<bits/stdc++.h>
+using namespace std;
+int getwater(int arr[],int n)
+{
+   int res=0;
+   int lmax[n],rmax[n];
+
+   	//left max
+   lmax[0]=arr[0];
+   for(int i=1;i<n;i++)
+        lmax[i]=max(arr[i],lmax[i-1]);
+
+			//starts from start to end
+			//1st element as it is other max will place in lmax
+			//Array is 5 0 6 2 3
+//					   |/\/\/\/\
+			 //lmax is 5 5 6 6 6
+
+	//right max
+   	rmax[n-1]=arr[n-1];
+   for(int i=n-2;i>=0;i--)
+    rmax[i]=max(arr[0],rmax[i+1]);
+
+	//starts from end to start
+	//last element as it is other max will place in rmax
+		//Array is 5 0 6 2 3
+//			       /\/\/\/\|
+   		 //rmax is 6 6 6 3 3
+
+   //here we get result
+   for(int i=1;i<n-1;i++)
+    res+=(min(lmax[i],rmax[i])-arr[i]);
+
+//    min(lmax,rmax)-arr[i]    min        arr[i]
+//		5 5 6 6 6			5 5 6 3 3 - 5 0 6 2 3 = 0+5+0+1+0=6
+//		6 6 6 3 3
+
+
+   return res;
+}
+int main()
+{
+	int n;
+    cout<<"Enter the number of elements in the array\n";
+    cin>>n;
+    int arr[n];
+    cout<<"Enter the elements in  the array\n";
+    for(int i=0;i<n;i++)
+        cin>>arr[i];
+    cout<<"Total trapped water is "<<getwater(arr,n);
+    return 0;
+}

Arrays/trapping_rain_water_two_pointer.cpp

@@ -0,0 +1,62 @@
+// Trapping Rain Water
+//Approach 3 (2 pointer approach)
+// O(n) O(1)
+
+//i/p: 3 0 1 2 5
+//o/p: 6
+
+/*
+We take 2 pointer left right
+and increment left and decrement right
+till we meet at a point
+*/
+
+
+#include<bits/stdc++.h>
+using namespace std;
+int getwater(int arr[],int n)
+{
+ 	int left=0,right=n-1;
+ 	int res=0;
+ 	int leftmax=0,rightmax=0;
+
+ 	// i/p: 7 1 4 0 2 8 6 3 5
+	//o/p: 23
+
+ 	while(left<=right)
+ 	{
+ 		//left side
+
+ 		if(arr[left]<=arr[right])
+		 {
+		 	if(arr[left]>=leftmax)
+		 		leftmax=arr[left];
+		 	else
+		 		res+=leftmax-arr[left]; // (7-1)=6 + (7-4)=3 + (7-0)=7 + (7-2)=5 ==> 21
+
+		 	left++; //increment
+		 }
+		 //right side
+		 	else
+		 	{
+		 		if(arr[right]>=rightmax)
+		 		    rightmax=arr[right]; //5
+		 		else
+		 		    res+=rightmax-arr[right]; //(5-3)=2 ==> 2
+		 		right--; //decrement
+		    }
+	}
+	return res;
+}
+int main()
+{
+	int n;
+    cout<<"Enter the number of elements in the array\n";
+    cin>>n;
+    int arr[n];
+    cout<<"Enter the elements in  the array\n";
+    for(int i=0;i<n;i++)
+        cin>>arr[i];
+    cout<<"Total trapped water is "<<getwater(arr,n);
+    return 0;
+}