Added tasks 3270-3277

Author: javadev

src/main/java/g1601_1700/s1616_split_two_strings_to_make_palindrome/Solution.java

@@ -1,28 +1,32 @@
 package g1601_1700.s1616_split_two_strings_to_make_palindrome;
 
-// #Medium #String #Greedy #Two_Pointers #2022_04_13_Time_4_ms_(89.77%)_Space_43.3_MB_(85.58%)
+// #Medium #String #Greedy #Two_Pointers #2024_09_04_Time_2_ms_(100.00%)_Space_45.1_MB_(97.99%)
 
 @SuppressWarnings("java:S2234")
 public class Solution {
     public boolean checkPalindromeFormation(String a, String b) {
-        return check(a, b) || check(b, a);
-    }
-
-    private boolean check(String a, String b) {
-        int i = 0;
-        int j = b.length() - 1;
-        while (j > i && a.charAt(i) == b.charAt(j)) {
-            ++i;
-            --j;
+        int n = a.length();
+        int s = 0;
+        int e = n - 1;
+        if (isPalindrome(a, b, s, e, true)) {
+            return true;
+        } else {
+            return isPalindrome(b, a, s, e, true);
         }
-        return isPalindrome(a, i, j) || isPalindrome(b, i, j);
     }
 
-    private boolean isPalindrome(String s, int i, int j) {
-        while (j > i && s.charAt(i) == s.charAt(j)) {
-            ++i;
-            --j;
+    private boolean isPalindrome(String a, String b, int s, int e, boolean check) {
+        if (s == e) {
+            return true;
+        }
+        while (s < e) {
+            if (a.charAt(s) != b.charAt(e)) {
+                return check
+                        && (isPalindrome(a, a, s, e, false) || isPalindrome(b, b, s, e, false));
+            }
+            s++;
+            e--;
         }
-        return i >= j;
+        return true;
     }
 }

src/main/java/g3201_3300/s3270_find_the_key_of_the_numbers/Solution.java

@@ -0,0 +1,18 @@
+package g3201_3300.s3270_find_the_key_of_the_numbers;
+
+// #Easy #Math #2024_09_02_Time_0_ms_(100.00%)_Space_40.5_MB_(100.00%)
+
+public class Solution {
+    public int generateKey(int num1, int num2, int num3) {
+        int s1 =
+                Math.min(((num1 / 1000) % 10), Math.min(((num2 / 1000) % 10), ((num3 / 1000) % 10)))
+                        * 1000;
+        int s2 =
+                Math.min(((num1 / 100) % 10), Math.min(((num2 / 100) % 10), ((num3 / 100) % 10)))
+                        * 100;
+        int s3 =
+                Math.min(((num1 / 10) % 10), Math.min(((num2 / 10) % 10), ((num3 / 10) % 10))) * 10;
+        int s4 = Math.min((num1 % 10), Math.min((num2 % 10), (num3 % 10)));
+        return s1 + s2 + s3 + s4;
+    }
+}

src/main/java/g3201_3300/s3270_find_the_key_of_the_numbers/readme.md

@@ -0,0 +1,45 @@
+3270\. Find the Key of the Numbers
+
+Easy
+
+You are given three **positive** integers `num1`, `num2`, and `num3`.
+
+The `key` of `num1`, `num2`, and `num3` is defined as a four-digit number such that:
+
+*   Initially, if any number has **less than** four digits, it is padded with **leading zeros**.
+*   The <code>i<sup>th</sup></code> digit (`1 <= i <= 4`) of the `key` is generated by taking the **smallest** digit among the <code>i<sup>th</sup></code> digits of `num1`, `num2`, and `num3`.
+
+Return the `key` of the three numbers **without** leading zeros (_if any_).
+
+**Example 1:**
+
+**Input:** num1 = 1, num2 = 10, num3 = 1000
+
+**Output:** 0
+
+**Explanation:**
+
+On padding, `num1` becomes `"0001"`, `num2` becomes `"0010"`, and `num3` remains `"1000"`.
+
+*   The <code>1<sup>st</sup></code> digit of the `key` is `min(0, 0, 1)`.
+*   The <code>2<sup>nd</sup></code> digit of the `key` is `min(0, 0, 0)`.
+*   The <code>3<sup>rd</sup></code> digit of the `key` is `min(0, 1, 0)`.
+*   The <code>4<sup>th</sup></code> digit of the `key` is `min(1, 0, 0)`.
+
+Hence, the `key` is `"0000"`, i.e. 0.
+
+**Example 2:**
+
+**Input:** num1 = 987, num2 = 879, num3 = 798
+
+**Output:** 777
+
+**Example 3:**
+
+**Input:** num1 = 1, num2 = 2, num3 = 3
+
+**Output:** 1
+
+**Constraints:**
+
+*   `1 <= num1, num2, num3 <= 9999`

src/main/java/g3201_3300/s3271_hash_divided_string/Solution.java

@@ -0,0 +1,20 @@
+package g3201_3300.s3271_hash_divided_string;
+
+// #Medium #String #Simulation #2024_09_02_Time_2_ms_(100.00%)_Space_44.7_MB_(100.00%)
+
+public class Solution {
+    public String stringHash(String s, int k) {
+        var result = new StringBuilder();
+        int i = 0;
+        int sum = 0;
+        while (i < s.length()) {
+            sum += s.charAt(i) - 'a';
+            if ((i + 1) % k == 0) {
+                result.append((char) ('a' + sum % 26));
+                sum = 0;
+            }
+            i++;
+        }
+        return result.toString();
+    }
+}

src/main/java/g3201_3300/s3271_hash_divided_string/readme.md

@@ -0,0 +1,46 @@
+3271\. Hash Divided String
+
+Medium
+
+You are given a string `s` of length `n` and an integer `k`, where `n` is a **multiple** of `k`. Your task is to hash the string `s` into a new string called `result`, which has a length of `n / k`.
+
+First, divide `s` into `n / k` **substrings**, each with a length of `k`. Then, initialize `result` as an **empty** string.
+
+For each **substring** in order from the beginning:
+
+*   The **hash value** of a character is the index of that character in the **English alphabet** (e.g., `'a' → 0`, `'b' → 1`, ..., `'z' → 25`).
+*   Calculate the _sum_ of all the **hash values** of the characters in the substring.
+*   Find the remainder of this sum when divided by 26, which is called `hashedChar`.
+*   Identify the character in the English lowercase alphabet that corresponds to `hashedChar`.
+*   Append that character to the end of `result`.
+
+Return `result`.
+
+**Example 1:**
+
+**Input:** s = "abcd", k = 2
+
+**Output:** "bf"
+
+**Explanation:**
+
+First substring: `"ab"`, `0 + 1 = 1`, `1 % 26 = 1`, `result[0] = 'b'`.
+
+Second substring: `"cd"`, `2 + 3 = 5`, `5 % 26 = 5`, `result[1] = 'f'`.
+
+**Example 2:**
+
+**Input:** s = "mxz", k = 3
+
+**Output:** "i"
+
+**Explanation:**
+
+The only substring: `"mxz"`, `12 + 23 + 25 = 60`, `60 % 26 = 8`, `result[0] = 'i'`.
+
+**Constraints:**
+
+*   `1 <= k <= 100`
+*   `k <= s.length <= 1000`
+*   `s.length` is divisible by `k`.
+*   `s` consists only of lowercase English letters.

src/main/java/g3201_3300/s3272_find_the_count_of_good_integers/Solution.java

@@ -0,0 +1,102 @@
+package g3201_3300.s3272_find_the_count_of_good_integers;
+
+// #Hard #Hash_Table #Math #Enumeration #Combinatorics
+// #2024_09_02_Time_167_ms_(100.00%)_Space_54.5_MB_(100.00%)
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.HashMap;
+import java.util.HashSet;
+import java.util.List;
+import java.util.Map;
+import java.util.Set;
+
+public class Solution {
+    private final List<String> palindromes = new ArrayList<>();
+
+    private long factorial(int n) {
+        long res = 1;
+        for (int i = 2; i <= n; i++) {
+            res *= i;
+        }
+        return res;
+    }
+
+    private Map<Character, Integer> countDigits(String s) {
+        Map<Character, Integer> freq = new HashMap<>();
+        for (char c : s.toCharArray()) {
+            freq.put(c, freq.getOrDefault(c, 0) + 1);
+        }
+        return freq;
+    }
+
+    private long calculatePermutations(Map<Character, Integer> freq, int length) {
+        long totalPermutations = factorial(length);
+        for (int count : freq.values()) {
+            totalPermutations /= factorial(count);
+        }
+        return totalPermutations;
+    }
+
+    private long calculateValidPermutations(String s) {
+        Map<Character, Integer> freq = countDigits(s);
+        int n = s.length();
+        long totalPermutations = calculatePermutations(freq, n);
+        if (freq.getOrDefault('0', 0) > 0) {
+            freq.put('0', freq.get('0') - 1);
+            long invalidPermutations = calculatePermutations(freq, n - 1);
+            totalPermutations -= invalidPermutations;
+        }
+        return totalPermutations;
+    }
+
+    private void generatePalindromes(
+            int f, int r, int k, int lb, int sum, StringBuilder ans, int[] rem) {
+        if (f > r) {
+            if (sum == 0) {
+                palindromes.add(ans.toString());
+            }
+            return;
+        }
+        for (int i = lb; i <= 9; i++) {
+            ans.setCharAt(f, (char) ('0' + i));
+            ans.setCharAt(r, (char) ('0' + i));
+            int chk = sum;
+            chk = (chk + rem[f] * i) % k;
+            if (f != r) {
+                chk = (chk + rem[r] * i) % k;
+            }
+            generatePalindromes(f + 1, r - 1, k, 0, chk, ans, rem);
+        }
+    }
+
+    private List<String> allKPalindromes(int n, int k) {
+        StringBuilder ans = new StringBuilder(n);
+        ans.append("0".repeat(Math.max(0, n)));
+        int[] rem = new int[n];
+        rem[0] = 1;
+        for (int i = 1; i < n; i++) {
+            rem[i] = (rem[i - 1] * 10) % k;
+        }
+        palindromes.clear();
+        generatePalindromes(0, n - 1, k, 1, 0, ans, rem);
+        return palindromes;
+    }
+
+    public long countGoodIntegers(int n, int k) {
+        List<String> ans = allKPalindromes(n, k);
+        Set<String> st = new HashSet<>();
+        for (String str : ans) {
+            char[] arr = str.toCharArray();
+            Arrays.sort(arr);
+            st.add(new String(arr));
+        }
+        List<String> v = new ArrayList<>(st);
+        long chk = 0;
+        for (String str : v) {
+            long cc = calculateValidPermutations(str);
+            chk += cc;
+        }
+        return chk;
+    }
+}

src/main/java/g3201_3300/s3272_find_the_count_of_good_integers/readme.md

@@ -0,0 +1,50 @@
+3272\. Find the Count of Good Integers
+
+Hard
+
+You are given two **positive** integers `n` and `k`.
+
+An integer `x` is called **k-palindromic** if:
+
+*   `x` is a palindrome.
+*   `x` is divisible by `k`.
+
+An integer is called **good** if its digits can be _rearranged_ to form a **k-palindromic** integer. For example, for `k = 2`, 2020 can be rearranged to form the _k-palindromic_ integer 2002, whereas 1010 cannot be rearranged to form a _k-palindromic_ integer.
+
+Return the count of **good** integers containing `n` digits.
+
+**Note** that _any_ integer must **not** have leading zeros, **neither** before **nor** after rearrangement. For example, 1010 _cannot_ be rearranged to form 101.
+
+**Example 1:**
+
+**Input:** n = 3, k = 5
+
+**Output:** 27
+
+**Explanation:**
+
+_Some_ of the good integers are:
+
+*   551 because it can be rearranged to form 515.
+*   525 because it is already k-palindromic.
+
+**Example 2:**
+
+**Input:** n = 1, k = 4
+
+**Output:** 2
+
+**Explanation:**
+
+The two good integers are 4 and 8.
+
+**Example 3:**
+
+**Input:** n = 5, k = 6
+
+**Output:** 2468
+
+**Constraints:**
+
+*   `1 <= n <= 10`
+*   `1 <= k <= 9`

src/main/java/g3201_3300/s3273_minimum_amount_of_damage_dealt_to_bob/Solution.java

@@ -0,0 +1,41 @@
+package g3201_3300.s3273_minimum_amount_of_damage_dealt_to_bob;
+
+// #Hard #Array #Sorting #Greedy #2024_09_04_Time_76_ms_(100.00%)_Space_59.5_MB_(61.02%)
+
+import java.util.Arrays;
+
+@SuppressWarnings("java:S1210")
+public class Solution {
+    public long minDamage(int pw, int[] damage, int[] health) {
+        long res = 0;
+        long sum = 0;
+        for (int e : damage) {
+            sum += e;
+        }
+        Pair[] pairs = new Pair[damage.length];
+        for (int e = 0; e < damage.length; e++) {
+            pairs[e] = new Pair(damage[e], (health[e] + pw - 1) / pw);
+        }
+        Arrays.sort(pairs);
+        for (Pair pr : pairs) {
+            res += pr.val * sum;
+            sum -= pr.key;
+        }
+        return res;
+    }
+
+    static class Pair implements Comparable<Pair> {
+        int key;
+        int val;
+
+        Pair(int key, int val) {
+            this.key = key;
+            this.val = val;
+        }
+
+        @Override
+        public int compareTo(Pair p) {
+            return val * p.key - key * p.val;
+        }
+    }
+}