# 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, 1, 6}. The largest possible rectangle possible is 12 (see the below figure, the max area rectangle is highlighted in red) ## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

## C++

 `// A Divide and Conquer Program to find maximum rectangular area in a histogram ` `#include ` `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); ` `    ``cout << ``"Maximum area is "` `<< getMaxArea(hist, n); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 program for range minimum   ` `# query using segment tree ` ` `  `# modified to return index of minimum instead of minimum itself ` `# for further reference link ` `# https://www.geeksforgeeks.org/segment-tree-set-1-range-minimum-query/ ` ` `  `#------------------------------------------------------------------------- ` `from` `math ``import` `ceil,log2;   ` `   `  `# A utility function to get   ` `# minimum of two numbers   ` `def` `minVal(hist,x, y) : ` `    ``if` `x``=``=``-``1``: ` `        ``return` `y ` `    ``if` `y``=``=``-``1``: ` `        ``return` `x ` `    ``return` `x ``if` `(hist[x] < hist[y]) ``else` `y;   ` `   `  `# A utility function to get the   ` `# middle index from corner indexes.   ` `def` `getMid(s, e) :  ` `    ``return` `s ``+` `(e ``-` `s) ``/``/` `2``;   ` `   `  `""" A recursive function to get the   ` `minimum value in a given range   ` `of array indexes. The following   ` `are parameters for this function.   ` `   `  `    ``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 """` `def` `RMQUtil( hist,st, ss, se, qs, qe, index) :  ` `   `  `    ``# If segment of this node is a part   ` `    ``# of given range, then return   ` `    ``# the min of the segment   ` `    ``if` `(qs <``=` `ss ``and` `qe >``=` `se) :  ` `        ``return` `st[index];   ` `   `  `    ``# If segment of this node   ` `    ``# is outside the given range   ` `    ``if` `(se < qs ``or` `ss > qe) :  ` `        ``return` `-``1``;   ` `   `  `    ``# If a part of this segment   ` `    ``# overlaps with the given range   ` `    ``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 minimum of elements in range   ` `# from index qs (query start) to   ` `# qe (query end). It mainly uses RMQUtil()   ` `def` `RMQ( hist,st, n, qs, qe) :   ` `   `  `    ``# Check for erroneous input values   ` `    ``if` `(qs < ``0` `or` `qe > n ``-` `1` `or` `qs > qe) :  ` `       `  `        ``print``(``"Invalid Input"``);   ` `        ``return` `-``1``;   ` `       `  `    ``return` `RMQUtil(hist,st, ``0``, n ``-` `1``, qs, qe, ``0``);   ` `   `  `# A recursive function that constructs   ` `# Segment Tree for array[ss..se].   ` `# si is index of current node in segment tree st   ` `def` `constructSTUtil(hist, ss, se, st, si) :  ` `   `  `    ``# If there is one element in array,   ` `    ``# store it in current node of   ` `    ``# segment tree and return   ` `    ``if` `(ss ``=``=` `se) :  ` `   `  `        ``st[si] ``=` `ss;   ` `        ``return` `st[si];   ` `   `  `    ``# If there are more than one elements,   ` `    ``# then recur for left and right subtrees   ` `    ``# and store the minimum of two values in this node   ` `    ``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 """` `def` `constructST( hist, n) :  ` `   `  `    ``# Allocate memory for segment tree   ` `   `  `    ``# Height of segment tree   ` `    ``x ``=` `(``int``)(ceil(log2(n)));   ` `   `  `    ``# Maximum size of segment tree   ` `    ``max_size ``=` `2` `*` `(``int``)(``2``*``*``x) ``-` `1``;   ` `    `  `    ``st ``=` `[``0``] ``*` `(max_size);   ` `   `  `    ``# Fill the allocated memory st   ` `    ``constructSTUtil(hist, ``0``, n ``-` `1``, st, ``0``);   ` `   `  `    ``# Return the constructed segment tree   ` `    ``return` `st;   ` `   `  ` `  `#---------------------------------------------------------------- ` ` `  `# main program ` `# Python3 program using Divide and Conquer ` `# to find maximum rectangular area under a histogram ` ` `  ` `  `def` `max_area_histogram(hist): ` `    ``area``=``0` `    ``#initialize area ` `     `  `    ``st ``=` `constructST(hist, ``len``(hist)) ` `    ``# contruct the segment tree ` `     `  `    ``try``: ` `        ``# try except block is generally used in this way ` `        ``# to suppress all type of exceptions raised. ` `         `  `        ``def` `fun(left,right): ` `             `  `        ``# this function "fun" calculates area ` `        ``# recursively between indices left and right ` `             `  `            ``nonlocal area ` `             `  `            ``# global area won't work here as ` `            ``# variable area is defined inside function ` `            ``# not in main(). ` `              `  `            ``if` `left``=``=``right: ` `                ``return` `            ``# the recursion has reached end ` `              `  `             `  `            ``index ``=` `RMQ(hist,st, ``len``(hist), left, right``-``1``) ` `            ``# RMQ function returns index  ` `            ``# of minimum value ` `            ``# in the range of [left,right-1] ` `            ``# can also be found by using min() but ` `            ``# results in O(n) instead of O(log n) for traversing ` `             `  `            ``area``=``max``(area,hist[index]``*``(right``-``left)) ` `            ``# calculate area with minimum above ` `             `  `            ``fun(index``+``1``,right) ` `            ``fun(left,index) ` `            ``# initiate further recursion ` `              `  `            ``return` `                  `  `        ``fun(``0``,``len``(hist)) ` `        ``# initializes the recursion ` `         `  `        ``return``(area) ` `        ``# return the max area to calling function ` `        ``# in this case "print" ` `          `  `    ``except``: ` `        ``pass` `     `  `# Driver Code  ` `hist ``=` `[``6``, ``2``, ``5``, ``4``, ``5``, ``1``, ``6``]  ` `print``(``"Maximum area is"``,   ` `       ``max_area_histogram(hist))  ` `   `  `# This code is contributed   ` `# by Vishnudev C. `

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