# Largest Rectangular Area in a Histogram | Set 1

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 same width and the width is 1 unit.

For example, consider the following histogram with 7 bars of heights {6, 2, 5, 4, 5, 2, 6}. The largest possible rectangle possible is 12 (see the below figure, the max area rectangle is highlighted in red)

A **simple solution** is to one by one consider all bars as starting points and calculate area of all rectangles starting with every bar. Finally return maximum of all possible areas. Time complexity of this solution would be O(n^2).

We can use **Divide and Conquer **to solve this in O(nLogn) time. The idea is to find the minimum value in the given array. Once we have index of the minimum value, the max area is maximum of following three values.

**a)** Maximum area in left side of minimum value (Not including the min value)

**b)** Maximum area in right side of minimum value (Not including the min value)

**c)** Number of bars multiplied by minimum value.

The areas in left and right of minimum value bar can be calculated recursively. If we use linear search to find the minimum value, then the worst case time complexity of this algorithm becomes O(n^2). In worst case, we always have (n-1) elements in one side and 0 elements in other side and if the finding minimum takes O(n) time, we get the recurrence similar to worst case of Quick Sort.

How to find the minimum efficiently? Range Minimum Query using Segment Tree can be used for this. We build segment tree of the given histogram heights. Once the segment tree is built, all range minimum queries take O(Logn) time. So over all complexity of the algorithm becomes.

Overall Time = Time to build Segment Tree + Time to recursively find maximum area

Time to build segment tree is O(n). Let the time to recursively find max area be T(n). It can be written as following.

T(n) = O(Logn) + T(n-1)

The solution of above recurrence is O(nLogn). So overall time is O(n) + O(nLogn) which is O(nLogn).

Following is C++ implementation of the above algorithm.

`// A Divide and Conquer Program to find maximum rectangular area in a histogram ` `#include <math.h> ` `#include <limits.h> ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// A utility function to find minimum of three integers ` `int` `max(` `int` `x, ` `int` `y, ` `int` `z) ` `{ ` `return` `max(max(x, y), z); } ` ` ` `// A utility function to get minimum of two numbers in hist[] ` `int` `minVal(` `int` `*hist, ` `int` `i, ` `int` `j) ` `{ ` ` ` `if` `(i == -1) ` `return` `j; ` ` ` `if` `(j == -1) ` `return` `i; ` ` ` `return` `(hist[i] < hist[j])? i : j; ` `} ` ` ` `// A utility function to get the middle index from corner indexes. ` `int` `getMid(` `int` `s, ` `int` `e) ` `{ ` `return` `s + (e -s)/2; } ` ` ` `/* A recursive function to get the index of minimum value in a given range of ` ` ` `indexes. The following are parameters for this function. ` ` ` ` ` `hist --> Input array for which segment tree is built ` ` ` `st --> Pointer to segment tree ` ` ` `index --> Index of current node in the segment tree. Initially 0 is ` ` ` `passed as root is always at index 0 ` ` ` `ss & se --> Starting and ending indexes of the segment represented by ` ` ` `current node, i.e., st[index] ` ` ` `qs & qe --> Starting and ending indexes of query range */` `int` `RMQUtil(` `int` `*hist, ` `int` `*st, ` `int` `ss, ` `int` `se, ` `int` `qs, ` `int` `qe, ` `int` `index) ` `{ ` ` ` `// If segment of this node is a part of given range, then return the ` ` ` `// min of the segment ` ` ` `if` `(qs <= ss && qe >= se) ` ` ` `return` `st[index]; ` ` ` ` ` `// If segment of this node is outside the given range ` ` ` `if` `(se < qs || ss > qe) ` ` ` `return` `-1; ` ` ` ` ` `// If a part of this segment overlaps with the given range ` ` ` `int` `mid = getMid(ss, se); ` ` ` `return` `minVal(hist, RMQUtil(hist, st, ss, mid, qs, qe, 2*index+1), ` ` ` `RMQUtil(hist, st, mid+1, se, qs, qe, 2*index+2)); ` `} ` ` ` `// Return index of minimum element in range from index qs (quey start) to ` `// qe (query end). It mainly uses RMQUtil() ` `int` `RMQ(` `int` `*hist, ` `int` `*st, ` `int` `n, ` `int` `qs, ` `int` `qe) ` `{ ` ` ` `// Check for erroneous input values ` ` ` `if` `(qs < 0 || qe > n-1 || qs > qe) ` ` ` `{ ` ` ` `cout << ` `"Invalid Input"` `; ` ` ` `return` `-1; ` ` ` `} ` ` ` ` ` `return` `RMQUtil(hist, st, 0, n-1, qs, qe, 0); ` `} ` ` ` `// A recursive function that constructs Segment Tree for hist[ss..se]. ` `// si is index of current node in segment tree st ` `int` `constructSTUtil(` `int` `hist[], ` `int` `ss, ` `int` `se, ` `int` `*st, ` `int` `si) ` `{ ` ` ` `// If there is one element in array, store it in current node of ` ` ` `// segment tree and return ` ` ` `if` `(ss == se) ` ` ` `return` `(st[si] = ss); ` ` ` ` ` `// If there are more than one elements, then recur for left and ` ` ` `// right subtrees and store the minimum of two values in this node ` ` ` `int` `mid = getMid(ss, se); ` ` ` `st[si] = minVal(hist, constructSTUtil(hist, ss, mid, st, si*2+1), ` ` ` `constructSTUtil(hist, mid+1, se, st, si*2+2)); ` ` ` `return` `st[si]; ` `} ` ` ` `/* Function to construct segment tree from given array. This function ` ` ` `allocates memory for segment tree and calls constructSTUtil() to ` ` ` `fill the allocated memory */` `int` `*constructST(` `int` `hist[], ` `int` `n) ` `{ ` ` ` `// Allocate memory for segment tree ` ` ` `int` `x = (` `int` `)(` `ceil` `(log2(n))); ` `//Height of segment tree ` ` ` `int` `max_size = 2*(` `int` `)` `pow` `(2, x) - 1; ` `//Maximum size of segment tree ` ` ` `int` `*st = ` `new` `int` `[max_size]; ` ` ` ` ` `// Fill the allocated memory st ` ` ` `constructSTUtil(hist, 0, n-1, st, 0); ` ` ` ` ` `// Return the constructed segment tree ` ` ` `return` `st; ` `} ` ` ` `// A recursive function to find the maximum rectangular area. ` `// It uses segment tree 'st' to find the minimum value in hist[l..r] ` `int` `getMaxAreaRec(` `int` `*hist, ` `int` `*st, ` `int` `n, ` `int` `l, ` `int` `r) ` `{ ` ` ` `// Base cases ` ` ` `if` `(l > r) ` `return` `INT_MIN; ` ` ` `if` `(l == r) ` `return` `hist[l]; ` ` ` ` ` `// Find index of the minimum value in given range ` ` ` `// This takes O(Logn)time ` ` ` `int` `m = RMQ(hist, st, n, l, r); ` ` ` ` ` `/* Return maximum of following three possible cases ` ` ` `a) Maximum area in Left of min value (not including the min) ` ` ` `a) Maximum area in right of min value (not including the min) ` ` ` `c) Maximum area including min */` ` ` `return` `max(getMaxAreaRec(hist, st, n, l, m-1), ` ` ` `getMaxAreaRec(hist, st, n, m+1, r), ` ` ` `(r-l+1)*(hist[m]) ); ` `} ` ` ` `// The main function to find max area ` `int` `getMaxArea(` `int` `hist[], ` `int` `n) ` `{ ` ` ` `// Build segment tree from given array. This takes ` ` ` `// O(n) time ` ` ` `int` `*st = constructST(hist, n); ` ` ` ` ` `// Use recursive utility function to find the ` ` ` `// maximum area ` ` ` `return` `getMaxAreaRec(hist, st, n, 0, n-1); ` `} ` ` ` `// Driver program to test above functions ` `int` `main() ` `{ ` ` ` `int` `hist[] = {6, 1, 5, 4, 5, 2, 6}; ` ` ` `int` `n = ` `sizeof` `(hist)/` `sizeof` `(hist[0]); ` ` ` `cout << ` `"Maximum area is "` `<< getMaxArea(hist, n); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

Maximum area is 12

This problem can be solved in linear time. See below set 2 for linear time solution.

Linear time solution for Largest Rectangular Area in a Histogram

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