// Source : https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/ // Author : Hao Chen // Date : 2021-03-27 /***************************************************************************************************** * * You are given an array rectangles where rectangles[i] = [li, wi] represents the i^th rectangle of * length li and width wi. * * You can cut the i^th rectangle to form a square with a side length of k if both k <= li and k <= * wi. For example, if you have a rectangle [4,6], you can cut it to get a square with a side length * of at most 4. * * Let maxLen be the side length of the largest square you can obtain from any of the given rectangles. * * Return the number of rectangles that can make a square with a side length of maxLen. * * Example 1: * * Input: rectangles = [[5,8],[3,9],[5,12],[16,5]] * Output: 3 * Explanation: The largest squares you can get from each rectangle are of lengths [5,3,5,5]. * The largest possible square is of length 5, and you can get it out of 3 rectangles. * * Example 2: * * Input: rectangles = [[2,3],[3,7],[4,3],[3,7]] * Output: 3 * * Constraints: * * 1 <= rectangles.length <= 1000 * rectangles[i].length == 2 * 1 <= li, wi <= 10^9 * li != wi ******************************************************************************************************/ class Solution { public: int countGoodRectangles(vector>& rectangles) { return countGoodRectangles2(rectangles); return countGoodRectangles1(rectangles); } int countGoodRectangles1(vector>& rectangles) { int maxLen = 0; for(auto& rect : rectangles) { int len = min(rect[0], rect[1]); maxLen = max(maxLen, len); } int cnt = 0; for(auto& rect : rectangles) { if (maxLen <= rect[0] && maxLen <= rect[1]) cnt++; } return cnt; } int countGoodRectangles2(vector>& rectangles) { int maxLen = 0; int cnt = 0; for(auto& rect : rectangles) { int len = min(rect[0], rect[1]); if (len > maxLen ) { cnt = 1; maxLen = len; }else if (len == maxLen ) { cnt++; } } return cnt; } };