Maximum Rectangular Area in a Histogram

Author: sachuverma

DSA Crack Sheet/README.md

@@ -155,6 +155,8 @@
 - [Inserting at the end of stack](https://stackoverflow.com/questions/45130465/inserting-at-the-end-of-stack "view topic")
 - [Reverse a stack using recursion](https://www.geeksforgeeks.org/reverse-a-stack-using-recursion/ "view topic")
 - [Sort a stack](https://practice.geeksforgeeks.org/problems/sort-a-stack/1# "view question") - [Cpp Solution](./solutions/Sort%20a%20stack.cpp)
+- [Merge Intervals](https://leetcode.com/problems/merge-intervals/ "view question") - [Cpp Solution](./solutions/Merge%20Intervals.cpp)
+- [Maximum Rectangular Area in a Histogram](https://practice.geeksforgeeks.org/problems/maximum-rectangular-area-in-a-histogram-1587115620/1 "view question") - [Cpp Solution](./solutions/Maximum%20Rectangular%20Area%20in%20a%20Histogram.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 ### Heap

DSA Crack Sheet/solutions/Maximum Rectangular Area in a Histogram.cpp

@@ -0,0 +1,65 @@
+/*
+Maximum Rectangular Area in a Histogram
+==========================================
+
+Find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of contiguous bars. For simplicity, assume that all bars have the same width and the width is 1 unit.
+
+Example 1:
+Input:
+N = 7
+arr[] = {6,2,5,4,5,1,6}
+Output: 12
+Explanation: 
+
+Example 2:
+Input:
+N = 8
+arr[] = {7 2 8 9 1 3 6 5}
+Output: 16
+Explanation: Maximum size of the histogram 
+will be 8  and there will be 2 consecutive 
+histogram. And hence the area of the 
+histogram will be 8x2 = 16.
+Your Task:
+The task is to complete the function getMaxArea() which takes the array arr[] and its size N as inputs and finds the largest rectangular area possible and returns the answer.
+
+Expected Time Complxity : O(N)
+Expected Auxilliary Space : O(N)
+
+Constraints:
+1 ≤ N ≤ 106
+1 ≤ arr[i] ≤ 1012
+*/
+
+long long getMaxArea(long long arr[], int n)
+{
+  stack<int> st;
+  vector<int> left(n, -1), right(n, n);
+  for (int i = 0; i < n; ++i)
+  {
+    while (st.size() && arr[st.top()] >= arr[i])
+      st.pop();
+    left[i] = (st.size() ? st.top() : -1);
+    st.push(i);
+  }
+
+  while (st.size())
+    st.pop();
+
+  for (int i = n - 1; i >= 0; --i)
+  {
+    while (st.size() && arr[st.top()] >= arr[i])
+      st.pop();
+    right[i] = (st.size() ? st.top() : n);
+    st.push(i);
+  }
+
+  long long ans = 0;
+  for (int i = 0; i < n; ++i)
+  {
+    long long width = right[i] - left[i] - 1;
+    ans = max(ans, arr[i] * width);
+  }
+
+  return ans;
+}