Maximum rectangle in a matrix
Given a rows x cols binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area.
class Solution {
public int maximalRectangle(char[][] matrix) {
if (matrix.length <= 0) return 0;
int n = matrix.length;
int m = matrix[0].length;
int[][] dp = new int[n][m];
int maxArea = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == 0)
dp[i][j] = matrix[i][j] == '1' ? 1 : 0;
else
dp[i][j] = matrix[i][j] == '1' ? (dp[i-1][j] + 1) : 0;
int min = dp[i][j];
for (int k = j; k >= 0; k--) {
if (min == 0) break;
if (dp[i][k] < min) min = dp[i][k];
maxArea = Math.max(maxArea, min * (j - k + 1));
}
}
}
return maxArea;
}
}Related Problems
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.
You are given n balloons, indexed from 0 to n - 1. Each balloon is painted with a number on it represented by an array nums. You are asked to burst all the balloons.
If you burst the ith balloon, you will get nums[i - 1] * nums[i] * nums[i + 1] coins. If i - 1 or i + 1 goes out of bounds of the array, then treat it as if there is a balloon with a 1 painted on it.
Return the maximum coins you can collect by bursting the balloons wisely.
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
You may assume that you have an infinite number of each kind of coin.
There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.
Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.