New Problem Solution -"Number Of Rectangles That Can Form The Largest Square"

README.md

@@ -52,6 +52,7 @@ LeetCode
 |1734|[Decode XORed Permutation](https://leetcode.com/problems/decode-xored-permutation/) | [C++](./algorithms/cpp/decodeXORedPermutation/DecodeXoredPermutation.cpp)|Medium|
 |1733|[Minimum Number of People to Teach](https://leetcode.com/problems/minimum-number-of-people-to-teach/) | [C++](./algorithms/cpp/minimumNumberOfPeopleToTeach/MinimumNumberOfPeopleToTeach.cpp)|Medium|
 |1732|[Find the Highest Altitude](https://leetcode.com/problems/find-the-highest-altitude/) | [C++](./algorithms/cpp/findTheHighestAltitude/FindTheHighestAltitude.cpp)|Easy|
+|1725|[Number Of Rectangles That Can Form The Largest Square](https://leetcode.com/problems/number-of-rectangles-that-can-form-the-largest-square/) | [C++](./algorithms/cpp/numberOfRectanglesThatCanFormTheLargestSquare/NumberOfRectanglesThatCanFormTheLargestSquare.cpp)|Easy|
 |1625|[Lexicographically Smallest String After Applying Operations](https://leetcode.com/problems/lexicographically-smallest-string-after-applying-operations/) | [C++](./algorithms/cpp/lexicographicallySmallestStringAfterApplyingOperations/LexicographicallySmallestStringAfterApplyingOperations.cpp)|Medium|
 |1624|[Largest Substring Between Two Equal Characters](https://leetcode.com/problems/largest-substring-between-two-equal-characters/) | [C++](./algorithms/cpp/largestSubstringBetweenTwoEqualCharacters/LargestSubstringBetweenTwoEqualCharacters.cpp)|Easy|
 |1605|[Find Valid Matrix Given Row and Column Sums](https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/) | [C++](./algorithms/cpp/FindValidMatrixGivenRowAndColumnSums/FindValidMatrixGivenRowAndColumnSums.cpp)|Medium|

algorithms/cpp/numberOfRectanglesThatCanFormTheLargestSquare/NumberOfRectanglesThatCanFormTheLargestSquare.cpp

@@ -0,0 +1,74 @@
+// 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<vector<int>>& rectangles) {
+        return countGoodRectangles2(rectangles);
+        return countGoodRectangles1(rectangles); 
+    }
+    
+    int countGoodRectangles1(vector<vector<int>>& 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<vector<int>>& 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;
+    }
+};