g0001_0100.s0085_maximal_rectangle.Solution Maven / Gradle / Ivy
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package g0001_0100.s0085_maximal_rectangle;
// #Hard #Array #Dynamic_Programming #Matrix #Stack #Monotonic_Stack
// #2022_06_20_Time_3_ms_(99.68%)_Space_46.4_MB_(89.77%)
public class Solution {
public int maximalRectangle(char[][] matrix) {
/*
* idea: using [LC84 Largest Rectangle in Histogram]. For each row of the matrix, construct
* the histogram based on the current row and the previous histogram (up to the previous
* row), then compute the largest rectangle area using LC84.
*/
int m = matrix.length;
int n;
if (m == 0 || (n = matrix[0].length) == 0) {
return 0;
}
int i;
int j;
int res = 0;
int[] heights = new int[n];
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
if (matrix[i][j] == '0') {
heights[j] = 0;
} else {
heights[j] += 1;
}
}
res = Math.max(res, largestRectangleArea(heights));
}
return res;
}
private int largestRectangleArea(int[] heights) {
/*
* idea: scan and store if a[i-1]<=a[i] (increasing), then as long as a[i]a[i], which
* is a[j]*(i-j). And meanwhile, all these bars (a[j]'s) are already done, and thus are
* throwable (using pop() with a stack).
*
* We can use an array nLeftGeq[] of size n to simulate a stack. nLeftGeq[i] = the number
* of elements to the left of [i] having value greater than or equal to a[i] (including a[i]
* itself). It is also the index difference between [i] and the next index on the top of the
* stack.
*/
int n = heights.length;
if (n == 0) {
return 0;
}
int[] nLeftGeq = new int[n];
// the number of elements to the left
// of [i] with value >= heights[i]
nLeftGeq[0] = 1;
// preIdx=the index of stack.peek(), res=max area so far
int preIdx = 0;
int res = 0;
for (int i = 1; i < n; i++) {
nLeftGeq[i] = 1;
// notice that preIdx = i - 1 = peek()
while (preIdx >= 0 && heights[i] < heights[preIdx]) {
res = Math.max(res, heights[preIdx] * (nLeftGeq[preIdx] + i - preIdx - 1));
// pop()
nLeftGeq[i] += nLeftGeq[preIdx];
// peek() current top
preIdx = preIdx - nLeftGeq[preIdx];
}
if (preIdx >= 0 && heights[i] == heights[preIdx]) {
// pop()
nLeftGeq[i] += nLeftGeq[preIdx];
}
// otherwise nothing to do
preIdx = i;
}
// compute the rest largest rectangle areas with (indices of) bases
// on stack
while (preIdx >= 0 && 0 < heights[preIdx]) {
res = Math.max(res, heights[preIdx] * (nLeftGeq[preIdx] + n - preIdx - 1));
// peek() current top
preIdx = preIdx - nLeftGeq[preIdx];
}
return res;
}
}