adding the k each chatacyer

src/Coding Questions/Leetcode/TakeKofEachCharacterFromLeftandRight.java

@@ -0,0 +1,105 @@
+/**
+ * 2516. Take K of Each Character From Left and Right
+Solved
+Medium
+Topics
+Companies
+Hint
+You are given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k. Each minute, you may take either the leftmost character of s, or the rightmost character of s.
+
+Return the minimum number of minutes needed for you to take at least k of each character, or return -1 if it is not possible to take k of each character.
+
+ 
+
+Example 1:
+
+Input: s = "aabaaaacaabc", k = 2
+Output: 8
+Explanation: 
+Take three characters from the left of s. You now have two 'a' characters, and one 'b' character.
+Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters.
+A total of 3 + 5 = 8 minutes is needed.
+It can be proven that 8 is the minimum number of minutes needed.
+Example 2:
+
+Input: s = "a", k = 1
+Output: -1
+Explanation: It is not possible to take one 'b' or 'c' so return -1.
+ 
+
+Constraints:
+
+1 <= s.length <= 105
+s consists of only the letters 'a', 'b', and 'c'.
+0 <= k <= s.length
+
+ */
+
+public class TakeKofEachCharacterFromLeftandRight {
+
+    public static void main(String[] args) {
+        TakeKofEachCharacterFromLeftandRight tke = new TakeKofEachCharacterFromLeftandRight();
+        System.out.println(tke.takeCharactersSolution("aabaaaacaabc", 2));
+        System.out.println(tke.takeCharactersBestSolution("aabaaaacaabc", 2));
+    }
+
+    public int takeCharactersSolution(String s, int k) {
+        int[] freq = new int[3];
+        int n = s.length();
+
+        for (char c : s.toCharArray()) {
+            freq[c - 'a']++;
+        }
+
+        if (freq[0] < k || freq[1] < k || freq[2] < k) {
+            return -1;
+        }
+
+        int[] curr = new int[3];
+        int maxLen = 0;
+        int left = 0;
+
+        for (int right = 0; right < n; right++) {
+            curr[s.charAt(right) - 'a']++;
+
+            while (left <= right && (curr[0] > freq[0] - k || 
+                   curr[1] > freq[1] - k || 
+                   curr[2] > freq[2] - k)) {
+                curr[s.charAt(left) - 'a']--;
+                left++;
+            }
+
+            maxLen = Math.max(maxLen, right - left + 1);
+        }
+
+        return n - maxLen;
+    }
+
+    public int takeCharactersBestSolution(String s, int k) {
+        int[] arr = new int[3];
+        char[] c = s.toCharArray();
+        int cur,len = c.length;
+        for(cur = 0; cur < len; cur++){
+            arr[c[cur] - 'a']++;
+            if(arr[0] >= k && arr[1] >= k && arr[2] >= k) break;
+        }
+        if(cur == len) return -1;
+        int count = cur + 1,min = count,end = len - 1;
+        while(cur >= 0){
+            if(arr[c[cur] - 'a'] == k){
+                while(c[cur] != c[end]){
+                    arr[c[end] - 'a']++;
+                    end--;
+                    count++;
+                }
+                end--;
+            }else{
+                arr[c[cur] - 'a']--;
+                count--;
+                min = Math.min(count, min);
+            }
+            cur--; 
+        }
+        return min;
+    }
+}