Added tasks 3300-3307

Author: javadev

src/main/kotlin/g3201_3300/s3300_minimum_element_after_replacement_with_digit_sum/Solution.kt

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+package g3201_3300.s3300_minimum_element_after_replacement_with_digit_sum
+
+// #Easy #Array #Math #2024_10_01_Time_153_ms_(100.00%)_Space_36.5_MB_(95.24%)
+
+import kotlin.math.min
+
+class Solution {
+    fun minElement(nums: IntArray): Int {
+        var min = Int.Companion.MAX_VALUE
+        for (x in nums) {
+            min = min(min, solve(x))
+        }
+        return min
+    }
+
+    private fun solve(x: Int): Int {
+        var x = x
+        var sum = 0
+        while (x != 0) {
+            sum += x % 10
+            x /= 10
+        }
+        return sum
+    }
+}

src/main/kotlin/g3201_3300/s3300_minimum_element_after_replacement_with_digit_sum/readme.md

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+3300\. Minimum Element After Replacement With Digit Sum
+
+Easy
+
+You are given an integer array `nums`.
+
+You replace each element in `nums` with the **sum** of its digits.
+
+Return the **minimum** element in `nums` after all replacements.
+
+**Example 1:**
+
+**Input:** nums = [10,12,13,14]
+
+**Output:** 1
+
+**Explanation:**
+
+`nums` becomes `[1, 3, 4, 5]` after all replacements, with minimum element 1.
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,4]
+
+**Output:** 1
+
+**Explanation:**
+
+`nums` becomes `[1, 2, 3, 4]` after all replacements, with minimum element 1.
+
+**Example 3:**
+
+**Input:** nums = [999,19,199]
+
+**Output:** 10
+
+**Explanation:**
+
+`nums` becomes `[27, 10, 19]` after all replacements, with minimum element 10.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   <code>1 <= nums[i] <= 10<sup>4</sup></code>

src/main/kotlin/g3301_3400/s3301_maximize_the_total_height_of_unique_towers/Solution.kt

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+package g3301_3400.s3301_maximize_the_total_height_of_unique_towers
+
+// #Medium #Array #Sorting #Greedy #2024_10_01_Time_761_ms_(87.50%)_Space_68.1_MB_(77.50%)
+
+class Solution {
+    fun maximumTotalSum(maximumHeight: IntArray): Long {
+        maximumHeight.sort()
+        var result = maximumHeight[maximumHeight.size - 1].toLong()
+        var previousHeight = maximumHeight[maximumHeight.size - 1].toLong()
+        for (i in maximumHeight.size - 2 downTo 0) {
+            if (previousHeight == 1L) {
+                return -1
+            }
+            val height = maximumHeight[i].toLong()
+            if (height >= previousHeight) {
+                result = result + previousHeight - 1
+                previousHeight = previousHeight - 1
+            } else {
+                result = result + height
+                previousHeight = height
+            }
+        }
+        return result
+    }
+}

src/main/kotlin/g3301_3400/s3301_maximize_the_total_height_of_unique_towers/readme.md

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+3301\. Maximize the Total Height of Unique Towers
+
+Medium
+
+You are given an array `maximumHeight`, where `maximumHeight[i]` denotes the **maximum** height the <code>i<sup>th</sup></code> tower can be assigned.
+
+Your task is to assign a height to each tower so that:
+
+1.  The height of the <code>i<sup>th</sup></code> tower is a positive integer and does not exceed `maximumHeight[i]`.
+2.  No two towers have the same height.
+
+Return the **maximum** possible total sum of the tower heights. If it's not possible to assign heights, return `-1`.
+
+**Example 1:**
+
+**Input:** maximumHeight = [2,3,4,3]
+
+**Output:** 10
+
+**Explanation:**
+
+We can assign heights in the following way: `[1, 2, 4, 3]`.
+
+**Example 2:**
+
+**Input:** maximumHeight = [15,10]
+
+**Output:** 25
+
+**Explanation:**
+
+We can assign heights in the following way: `[15, 10]`.
+
+**Example 3:**
+
+**Input:** maximumHeight = [2,2,1]
+
+**Output:** \-1
+
+**Explanation:**
+
+It's impossible to assign positive heights to each index so that no two towers have the same height.
+
+**Constraints:**
+
+*   <code>1 <= maximumHeight.length <= 10<sup>5</sup></code>
+*   <code>1 <= maximumHeight[i] <= 10<sup>9</sup></code>

src/main/kotlin/g3301_3400/s3302_find_the_lexicographically_smallest_valid_sequence/Solution.kt

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+package g3301_3400.s3302_find_the_lexicographically_smallest_valid_sequence
+
+// #Medium #String #Dynamic_Programming #Greedy #Two_Pointers
+// #2024_10_01_Time_705_ms_(100.00%)_Space_65.9_MB_(100.00%)
+
+class Solution {
+    fun validSequence(word1: String, word2: String): IntArray? {
+        val c1 = word1.toCharArray()
+        val c2 = word2.toCharArray()
+        val dp = IntArray(c1.size + 1)
+        var j = c2.size - 1
+        for (i in c1.indices.reversed()) {
+            if (j >= 0 && c1[i] == c2[j]) {
+                dp[i] = dp[i + 1] + 1
+                j--
+            } else {
+                dp[i] = dp[i + 1]
+            }
+        }
+        val ans = IntArray(c2.size)
+        var i = 0
+        j = 0
+        while (i < c1.size && j < c2.size) {
+            if (c1[i] == c2[j]) {
+                ans[j] = i
+                j++
+            } else {
+                if (dp[i + 1] >= c2.size - 1 - j) {
+                    ans[j] = i
+                    j++
+                    i++
+                    break
+                }
+            }
+            i++
+        }
+        if (j < c2.size && i == c1.size) {
+            return IntArray(0)
+        }
+        while (j < c2.size && i < c1.size) {
+            if (c2[j] == c1[i]) {
+                ans[j] = i
+                j++
+            }
+            i++
+        }
+        return ans
+    }
+}

src/main/kotlin/g3301_3400/s3302_find_the_lexicographically_smallest_valid_sequence/readme.md

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+3302\. Find the Lexicographically Smallest Valid Sequence
+
+Medium
+
+You are given two strings `word1` and `word2`.
+
+A string `x` is called **almost equal** to `y` if you can change **at most** one character in `x` to make it _identical_ to `y`.
+
+A sequence of indices `seq` is called **valid** if:
+
+*   The indices are sorted in **ascending** order.
+*   _Concatenating_ the characters at these indices in `word1` in **the same** order results in a string that is **almost equal** to `word2`.
+
+Return an array of size `word2.length` representing the lexicographically smallest **valid** sequence of indices. If no such sequence of indices exists, return an **empty** array.
+
+**Note** that the answer must represent the _lexicographically smallest array_, **not** the corresponding string formed by those indices.
+
+**Example 1:**
+
+**Input:** word1 = "vbcca", word2 = "abc"
+
+**Output:** [0,1,2]
+
+**Explanation:**
+
+The lexicographically smallest valid sequence of indices is `[0, 1, 2]`:
+
+*   Change `word1[0]` to `'a'`.
+*   `word1[1]` is already `'b'`.
+*   `word1[2]` is already `'c'`.
+
+**Example 2:**
+
+**Input:** word1 = "bacdc", word2 = "abc"
+
+**Output:** [1,2,4]
+
+**Explanation:**
+
+The lexicographically smallest valid sequence of indices is `[1, 2, 4]`:
+
+*   `word1[1]` is already `'a'`.
+*   Change `word1[2]` to `'b'`.
+*   `word1[4]` is already `'c'`.
+
+**Example 3:**
+
+**Input:** word1 = "aaaaaa", word2 = "aaabc"
+
+**Output:** []
+
+**Explanation:**
+
+There is no valid sequence of indices.
+
+**Example 4:**
+
+**Input:** word1 = "abc", word2 = "ab"
+
+**Output:** [0,1]
+
+**Constraints:**
+
+*   <code>1 <= word2.length < word1.length <= 3 * 10<sup>5</sup></code>
+*   `word1` and `word2` consist only of lowercase English letters.

src/main/kotlin/g3301_3400/s3303_find_the_occurrence_of_first_almost_equal_substring/Solution.kt

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+package g3301_3400.s3303_find_the_occurrence_of_first_almost_equal_substring
+
+// #Hard #String #String_Matching #2024_10_01_Time_364_ms_(100.00%)_Space_40.8_MB_(100.00%)
+
+import kotlin.math.abs
+
+class Solution {
+    fun minStartingIndex(s: String, pattern: String): Int {
+        val n = s.length
+        var left = 0
+        var right = 0
+        val f1 = IntArray(26)
+        val f2 = IntArray(26)
+        for (ch in pattern.toCharArray()) {
+            f2[ch.code - 'a'.code]++
+        }
+        while (right < n) {
+            val ch = s[right]
+            f1[ch.code - 'a'.code]++
+            if (right - left + 1 == pattern.length + 1) {
+                f1[s[left].code - 'a'.code]--
+                left += 1
+            }
+            if (right - left + 1 == pattern.length && check(f1, f2, left, s, pattern)) {
+                return left
+            }
+            right += 1
+        }
+        return -1
+    }
+
+    private fun check(f1: IntArray, f2: IntArray, left: Int, s: String, pattern: String): Boolean {
+        var cnt = 0
+        for (i in 0..25) {
+            if (f1[i] != f2[i]) {
+                if ((abs((f1[i] - f2[i])) > 1) || (abs(f1[i] - f2[i]) != 1 && cnt == 2)) {
+                    return false
+                }
+                cnt += 1
+            }
+        }
+        cnt = 0
+        var start = 0
+        var end = pattern.length - 1
+        while (start <= end) {
+            if (s[start + left] != pattern[start]) {
+                cnt += 1
+            }
+            if (start + left != left + end && s[left + end] != pattern[end]) {
+                cnt += 1
+            }
+            if (cnt >= 2) {
+                return false
+            }
+            start++
+            end--
+        }
+        return true
+    }
+}

src/main/kotlin/g3301_3400/s3303_find_the_occurrence_of_first_almost_equal_substring/readme.md

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+3303\. Find the Occurrence of First Almost Equal Substring
+
+Hard
+
+You are given two strings `s` and `pattern`.
+
+A string `x` is called **almost equal** to `y` if you can change **at most** one character in `x` to make it _identical_ to `y`.
+
+Return the **smallest** _starting index_ of a substring in `s` that is **almost equal** to `pattern`. If no such index exists, return `-1`.
+
+A **substring** is a contiguous **non-empty** sequence of characters within a string.
+
+**Example 1:**
+
+**Input:** s = "abcdefg", pattern = "bcdffg"
+
+**Output:** 1
+
+**Explanation:**
+
+The substring `s[1..6] == "bcdefg"` can be converted to `"bcdffg"` by changing `s[4]` to `"f"`.
+
+**Example 2:**
+
+**Input:** s = "ababbababa", pattern = "bacaba"
+
+**Output:** 4
+
+**Explanation:**
+
+The substring `s[4..9] == "bababa"` can be converted to `"bacaba"` by changing `s[6]` to `"c"`.
+
+**Example 3:**
+
+**Input:** s = "abcd", pattern = "dba"
+
+**Output:** \-1
+
+**Example 4:**
+
+**Input:** s = "dde", pattern = "d"
+
+**Output:** 0
+
+**Constraints:**
+
+*   <code>1 <= pattern.length < s.length <= 3 * 10<sup>5</sup></code>
+*   `s` and `pattern` consist only of lowercase English letters.
+
+**Follow-up:** Could you solve the problem if **at most** `k` **consecutive** characters can be changed?