update: weekly contest, pass 3 problems

daily problem

daily/May26.cpp

@@ -0,0 +1,124 @@
+#include <bits/stdc++.h>
+#include <bits/types.h>
+
+#include <algorithm>
+#include <unordered_map>
+
+using namespace std;
+
+class Solution {
+ public:
+  int numberOfPairs(vector<int>& nums1, vector<int>& nums2, int k) {
+    for (int i = 0; i < nums2.size(); i++) {
+      nums2[i] *= k;
+    }
+
+    int res = 0;
+    for (int i = 0; i < nums1.size(); i++) {
+      for (int j = 0; j < nums2.size(); j++) {
+        if (nums1[i] >= nums2[j] && nums1[i] % nums2[j] == 0) {
+          res++;
+        }
+      }
+    }
+    return res;
+  }
+
+  string compressedString(string word) {
+    if (word.empty()) {
+      return word;
+    }
+
+    // Initialize the compressed string
+    std::string compressed;
+    // Initialize the count of the current character
+    int count = 1;
+
+    // Iterate through the string starting from the second character
+    for (size_t i = 1; i < word.length(); ++i) {
+      // Check if the current character is the same as the previous one
+      if (word[i] == word[i - 1]) {
+        // Increment the count of the current character
+        count++;
+      } else {
+        // Append the previous character and its count(s) to the compressed
+        // string
+        while (count > 9) {
+          compressed += '9';
+          compressed += word[i - 1];
+          count -= 9;
+        }
+        compressed += word[i - 1];
+        compressed += std::to_string(count);
+        // Reset the count for the new character
+        count = 1;
+      }
+    }
+
+    // Append the last character and its count(s)
+    while (count > 9) {
+      compressed += '9';
+      compressed += word.back();
+      count -= 9;
+    }
+    compressed += std::to_string(count);
+    compressed += word.back();
+
+    // Return the original string if the compressed string is not shorter
+    // return compressed.length() < word.length() ? compressed : word;
+    return compressed;
+  }
+
+  long long numberOfPairsII(vector<int>& nums1, vector<int>& nums2, int k) {
+    unordered_map<int, int> map;
+    long long pairs = 0;
+
+    // Count occurrences of elements in nums2
+    for (int num : nums2) {
+      map[num]++;
+    }
+
+    // Iterate through nums1 and check for divisible pairs
+    for (int num : nums1) {
+      if (num % k == 0) {
+        int divisor = num / k;
+
+        // Check for divisors up to the square root of divisor
+        for (int j = 1; j * j <= divisor; j++) {
+          if (divisor % j == 0) {
+            pairs += map[j];
+            if (j != divisor / j) {
+              pairs += map[divisor / j];
+            }
+          }
+        }
+      }
+    }
+
+    return pairs;
+  }
+
+  /**
+   * @brief LC: 1738: Find Kth Largest XOR Coordinate Val
+   * Time: O(mnlog(mn))  / O(mn)
+   * Space: O(MN)
+   *
+   * @param matrix
+   * @param k
+   * @return int
+   */
+  int kthLargestValue(vector<vector<int>>& matrix, int k) {
+    int m = matrix.size(), n = matrix[0].size();
+    vector<int> a;
+    vector<vector<int>> s(m + 1, vector<int>(n + 1));
+    for (int i = 0; i < m; i++) {
+      for (int j = 0; j < n; j++) {
+        s[i + 1][j + 1] = s[i + 1][j] ^ s[i][j + 1] ^ s[i][j] ^ matrix[i][j];
+      }
+      a.insert(a.end(), s[i + 1].begin() + 1, s[i + 1].end());
+    }
+    // ranges::nth_element(a, a.end() - k);
+    std::nth_element(a.begin(), a.begin() + k, a.end(), std::greater<int>());
+    return a[a.size() - k];
+  }
+};