Related: LeetCode 84 - Largest Rectangle in Histogram
LeetCode 84 - Largest Rectangle in 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 same width and the width is 1 unit.
LeetCode 84 - Largest Rectangle in 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 same width and the width is 1 unit.
2) Start from first bar, and do following for every bar ‘hist[i]‘ where ‘i’ varies from 0 to n-1.
……a) If stack is empty or hist[i] is higher than the bar at top of stack, then push ‘i’ to stack.
……b) If this bar is smaller than the top of stack, then keep removing the top of stack while top of the stack is greater. Let the removed bar be hist[tp]. Calculate area of rectangle with hist[tp] as smallest bar. For hist[tp], the ‘left index’ is previous (previous to tp) item in stack and ‘right index’ is ‘i’ (current index).
3) If the stack is not empty, then one by one remove all bars from stack and do step 2.b for every removed bar.
For every bar ‘x’, we calculate the area with ‘x’ as the smallest bar in the rectangle. If we calculate such area for every bar ‘x’ and find the maximum of all areas, our task is done.
We have a stack, and push into it the indices of the bars one by one. But before the push, we pop out the indices of the bars (if any) that are higher than the one under investigation. Upon each pop, we calculate the area of largest rectangle containing the popped bar as follows: if the stack is empty, it means all previous bars are all higher than it, so the area to be calculated would be the product of its height and the number of the bars including and before it; otherwise, it means the heights of the bars between the one in the top of the stack and the one under investigation (both exclusive) are no smaller than that of the popped one, so the area would be the product of its height and the number of those bars.
X. Using stack
https://discuss.leetcode.com/topic/7599/o-n-stack-based-java-solution
- Do push all heights including 0 height.
i - 1 - s.peek()uses the starting index whereheight[s.peek() + 1] >= height[tp], because the index on top of the stack right now is the first index left oftpwith height smaller than tp's height.
public int largestRectangleArea(int[] height) {
int len = height.length;
Stack<Integer> s = new Stack<Integer>();
int maxArea = 0;
for(int i = 0; i <= len; i++){
int h = (i == len ? 0 : height[i]);
if(s.isEmpty() || h >= height[s.peek()]){
s.push(i);
}else{
int tp = s.pop();
maxArea = Math.max(maxArea, height[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
i--;
}
}
return maxArea;
}
public int largestRectangleArea(int[] height) {
if (height == null || height.length == 0)
return 0;
int n = height.length;
int largestArea = 0;
Deque<Integer> indexStack = new LinkedList<Integer>();
// For each bar, pop out the bars higher than it and calculate the largest area of
// the rectangle using this full bar, before pushing it to the stack
for (int i = 0; i < n; i++) {
while (!indexStack.isEmpty() && height[i] <= height[indexStack.peek()]) {
int index = indexStack.pop();
if (indexStack.isEmpty())
largestArea = Math.max(largestArea, i*height[index]);
else
largestArea = Math.max(largestArea, (i-indexStack.peek()-1)*height[index]);
}
indexStack.push(i);
}
// Calculate the largest area of the rectangle using each full bar remaining in the stack
while (!indexStack.isEmpty()) {
int index = indexStack.pop();
if (indexStack.isEmpty())
largestArea = Math.max(largestArea, n*height[index]);
else
largestArea = Math.max(largestArea, (n-indexStack.peek()-1)*height[index]);
}
return largestArea;
}
http://dp2.me/blog/?p=482
在这个解法中，我们首先计算一个高度向左向右最大可以延伸的范围，这个量我们简称为“左/右可达”。具体来说，假设我们已经解决了k – 1规模的问题，当一个新点到达时，我们便向左看，如果左边的值比它小 ，那么它本身的“左可达”，即是它自身；若左边的值比它大，那么它本身的“左可达”，至少是和左边的值一样——于是我们跳过一段（因为我们已经有了k-1规模的解），再去看这一段左边的高度是否比当前高度小，然后重复以上逻辑。注意到这里的状态是不断被压缩的——当我们发现新到来的高度比左边的高度小的时候，至少左边的这一个高度再也不会被访问到。
最后，我们需要证明这个算法是O(n)的。这里我们需要看到，在比较过程中，我们已经处理过的高度比新高度大，对于这个已经处理过的高度，这种情况最多出现一次，以后我们再也不会做这个比较——因为新来的点的左可达已经延伸到更左边的地方，且这个新来的高度比已经处理过的高度小——再考虑一下后继比较即可；而已经处理过的高度比新高度小，这种情况每次比较都会出现正好一次，导致比较中止。所以整体复杂度是线性的。
int largestRectangleArea(vector<int> &height) {
if (height.size() == 0) {
return 0;
}
int maxArea = 0;
height.push_back(-1);
stack<int> expandLeft;
expandLeft.push(0);
for (int i = 1; i < height.size(); ++i) {
int j = i;
while(j > 0) {
if (height[j - 1] < height[i]) {
break;
} else {
maxArea = max(maxArea, (i - expandLeft.top()) * height[j - 1]);
j = expandLeft.top();
expandLeft.pop();
}
}
expandLeft.push(j);
}
height.pop_back();
return maxArea;
}
http://www.geeksforgeeks.org/largest-rectangle-under-histogram/
// The main function to find the maximum rectangular area under given// histogram with n barsint getMaxArea(int hist[], int n){ // Create an empty stack. The stack holds indexes of hist[] array // The bars stored in stack are always in increasing order of their // heights. stack<int> s; int max_area = 0; // Initalize max area int tp; // To store top of stack int area_with_top; // To store area with top bar as the smallest bar // Run through all bars of given histogram int i = 0; while (i < n) { // If this bar is higher than the bar on top stack, push it to stack if (s.empty() || hist[s.top()] <= hist[i]) s.push(i++); // If this bar is lower than top of stack, then calculate area of rectangle // with stack top as the smallest (or minimum height) bar. 'i' is // 'right index' for the top and element before top in stack is 'left index' else { tp = s.top(); // store the top index s.pop(); // pop the top // Calculate the area with hist[tp] stack as smallest bar area_with_top = hist[tp] * (s.empty() ? i : i - s.top() - 1); // update max area, if needed if (max_area < area_with_top) max_area = area_with_top; } } // Now pop the remaining bars from stack and calculate area with every // popped bar as the smallest bar while (s.empty() == false) { tp = s.top(); s.pop(); area_with_top = hist[tp] * (s.empty() ? i : i - s.top() - 1); if (max_area < area_with_top) max_area = area_with_top; } return max_area;}http://www.geeksforgeeks.org/largest-rectangular-area-in-a-histogram-set-1/
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.
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).
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).
/* 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 stint 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 areaint 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);}Read full article from Largest Rectangular Area in a Histogram | Set 2 | GeeksforGeeks
