Completed Hashing-2

contiguosarray.cpp

@@ -0,0 +1,28 @@
+class Solution {
+public:
+    int findMaxLength(vector<int>& nums) {
+        int n = nums.size();
+        unordered_map<int , int> map;
+        map[0] = -1;//added in order to handle subarrays that start from index 0
+        int rSum = 0;//aka prefix sum
+
+        int max = 0;//stores the maximum length of a balanced subarray found so far
+
+        for (int i = 0; i < n; i++) {
+            int num = nums[i];
+
+            if (num == 0) rSum--;//-1 if 0
+            else rSum++;//+1 if 1
+
+            if (map.find(rSum) != map.end()){//search if rSum was seen in the hashmap
+                max = std::max(max, i - map[rSum]);//computes the length of the max arraay length so far
+            }
+            else{
+                map[rSum]= i;//rSum was not seen before
+
+            }
+        }
+        return max;
+    }
+
+};

longestpalindrome.cpp

@@ -0,0 +1,35 @@
+class Solution {
+public:
+    int longestPalindrome(string s) {
+        unordered_map<char, int> freq;
+
+        // count frequency of each character in the string
+        for (char c : s) {
+            freq[c]++;   // increment count of character c
+        }
+        int length = 0;      
+        bool hasOdd = false; // flag to check if any character has odd frequency
+
+        // iterate over each character frequency
+        for (auto it : freq) {
+            // it.second -> frequency of that character
+
+            // if frequency is even, we can use all characters
+            if (it.second % 2 == 0) {
+                length += it.second;
+            }
+            else {
+                length += it.second - 1;
+
+                // mark that at least one odd-frequency character exists
+                hasOdd = true;
+            }
+        }
+
+        // we can place exactly one character in the center of the palindrome
+        if (hasOdd) {
+            length += 1;
+        }
+        return length;
+    }
+};