Added tasks 2914-2929, 2945-2949

Author: javadev

src/main/kotlin/g2901_3000/s2914_minimum_number_of_changes_to_make_binary_string_beautiful/Solution.kt

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+package g2901_3000.s2914_minimum_number_of_changes_to_make_binary_string_beautiful
+
+// #Medium #String #2023_12_31_Time_180_ms_(100.00%)_Space_38.2_MB_(80.00%)
+
+class Solution {
+    fun minChanges(s: String): Int {
+        var ans = 0
+        var i = 0
+        while (i < s.length) {
+            if (s[i] != s[i + 1]) {
+                ans++
+            }
+            i += 2
+        }
+        return ans
+    }
+}

src/main/kotlin/g2901_3000/s2914_minimum_number_of_changes_to_make_binary_string_beautiful/readme.md

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+2914\. Minimum Number of Changes to Make Binary String Beautiful
+
+Medium
+
+You are given a **0-indexed** binary string `s` having an even length.
+
+A string is **beautiful** if it's possible to partition it into one or more substrings such that:
+
+*   Each substring has an **even length**.
+*   Each substring contains **only** `1`'s or **only** `0`'s.
+
+You can change any character in `s` to `0` or `1`.
+
+Return _the **minimum** number of changes required to make the string_ `s` _beautiful_.
+
+**Example 1:**
+
+**Input:** s = "1001"
+
+**Output:** 2
+
+**Explanation:** We change s[1] to 1 and s[3] to 0 to get string "1100". It can be seen that the string "1100" is beautiful because we can partition it into "11|00". It can be proven that 2 is the minimum number of changes needed to make the string beautiful.
+
+**Example 2:**
+
+**Input:** s = "10"
+
+**Output:** 1
+
+**Explanation:** We change s[1] to 1 to get string "11". It can be seen that the string "11" is beautiful because we can partition it into "11". It can be proven that 1 is the minimum number of changes needed to make the string beautiful.
+
+**Example 3:**
+
+**Input:** s = "0000"
+
+**Output:** 0
+
+**Explanation:** We don't need to make any changes as the string "0000" is beautiful already.
+
+**Constraints:**
+
+*   <code>2 <= s.length <= 10<sup>5</sup></code>
+*   `s` has an even length.
+*   `s[i]` is either `'0'` or `'1'`.

src/main/kotlin/g2901_3000/s2915_length_of_the_longest_subsequence_that_sums_to_target/Solution.kt

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+package g2901_3000.s2915_length_of_the_longest_subsequence_that_sums_to_target
+
+// #Medium #Array #Dynamic_Programming #2023_12_31_Time_552_ms_(66.67%)_Space_39.4_MB_(83.33%)
+
+import kotlin.math.max
+
+class Solution {
+    fun lengthOfLongestSubsequence(nums: List<Int>, target: Int): Int {
+        val dp = IntArray(target + 1)
+        for (i in 1..target) {
+            dp[i] = -1
+        }
+        dp[0] = 0
+        for (num in nums) {
+            for (j in target downTo num) {
+                if (dp[j - num] != -1) {
+                    dp[j] = max(dp[j], dp[j - num] + 1)
+                }
+            }
+        }
+        if (dp[target] == -1) {
+            return -1
+        }
+        return dp[target]
+    }
+}

src/main/kotlin/g2901_3000/s2915_length_of_the_longest_subsequence_that_sums_to_target/readme.md

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+2915\. Length of the Longest Subsequence That Sums to Target
+
+Medium
+
+You are given a **0-indexed** array of integers `nums`, and an integer `target`.
+
+Return _the **length of the longest subsequence** of_ `nums` _that sums up to_ `target`. _If no such subsequence exists, return_ `-1`.
+
+A **subsequence** is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,4,5], target = 9
+
+**Output:** 3
+
+**Explanation:** There are 3 subsequences with a sum equal to 9: [4,5], [1,3,5], and [2,3,4]. The longest subsequences are [1,3,5], and [2,3,4]. Hence, the answer is 3.
+
+**Example 2:**
+
+**Input:** nums = [4,1,3,2,1,5], target = 7
+
+**Output:** 4
+
+**Explanation:** There are 5 subsequences with a sum equal to 7: [4,3], [4,1,2], [4,2,1], [1,1,5], and [1,3,2,1]. The longest subsequence is [1,3,2,1]. Hence, the answer is 4.
+
+**Example 3:**
+
+**Input:** nums = [1,1,5,4,5], target = 3
+
+**Output:** -1
+
+**Explanation:** It can be shown that nums has no subsequence that sums up to 3.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 1000`
+*   `1 <= nums[i] <= 1000`
+*   `1 <= target <= 1000`

src/main/kotlin/g2901_3000/s2916_subarrays_distinct_element_sum_of_squares_ii/Solution.kt

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+package g2901_3000.s2916_subarrays_distinct_element_sum_of_squares_ii
+
+// #Hard #Array #Dynamic_Programming #Segment_Tree #Binary_Indexed_Tree
+// #2023_12_31_Time_467_ms_(100.00%)_Space_58_MB_(100.00%)
+
+@Suppress("NAME_SHADOWING")
+class Solution {
+    private var n = 0
+    private lateinit var tree1: LongArray
+    private lateinit var tree2: LongArray
+
+    fun sumCounts(nums: IntArray): Int {
+        n = nums.size
+        tree1 = LongArray(n + 1)
+        tree2 = LongArray(n + 1)
+        var max = 0
+        for (x in nums) {
+            if (x > max) {
+                max = x
+            }
+        }
+        val last = IntArray(max + 1)
+        var ans: Long = 0
+        var cur: Long = 0
+        for (i in 1..n) {
+            val x = nums[i - 1]
+            val j = last[x]
+            cur += 2 * (query(i) - query(j)) + (i - j)
+            ans += cur
+            update(j + 1, 1)
+            update(i + 1, -1)
+            last[x] = i
+        }
+        return (ans % MOD).toInt()
+    }
+
+    private fun lowbit(index: Int): Int {
+        return index and (-index)
+    }
+
+    private fun update(index: Int, x: Int) {
+        var index = index
+        val v = index * x
+        while (index <= n) {
+            tree1[index] += x.toLong()
+            tree2[index] += v.toLong()
+            index += lowbit(index)
+        }
+    }
+
+    private fun query(index: Int): Long {
+        var index = index
+        var res: Long = 0
+        val p = index + 1
+        while (index > 0) {
+            res += p * tree1[index] - tree2[index]
+            index -= lowbit(index)
+        }
+        return res
+    }
+
+    companion object {
+        private const val MOD = 1e9.toInt() + 7
+    }
+}

src/main/kotlin/g2901_3000/s2916_subarrays_distinct_element_sum_of_squares_ii/readme.md

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+2916\. Subarrays Distinct Element Sum of Squares II
+
+Hard
+
+You are given a **0-indexed** integer array `nums`.
+
+The **distinct count** of a subarray of `nums` is defined as:
+
+*   Let `nums[i..j]` be a subarray of `nums` consisting of all the indices from `i` to `j` such that `0 <= i <= j < nums.length`. Then the number of distinct values in `nums[i..j]` is called the distinct count of `nums[i..j]`.
+
+Return _the sum of the **squares** of **distinct counts** of all subarrays of_ `nums`.
+
+Since the answer may be very large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+A subarray is a contiguous **non-empty** sequence of elements within an array.
+
+**Example 1:**
+
+**Input:** nums = [1,2,1]
+
+**Output:** 15
+
+**Explanation:** Six possible subarrays are: 
+
+[1]: 1 distinct value 
+
+[2]: 1 distinct value 
+
+[1]: 1 distinct value 
+
+[1,2]: 2 distinct values
+
+[2,1]: 2 distinct values
+
+[1,2,1]: 2 distinct values
+
+The sum of the squares of the distinct counts in all subarrays is equal to 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> = 15.
+
+**Example 2:**
+
+**Input:** nums = [2,2]
+
+**Output:** 3
+
+**Explanation:** Three possible subarrays are: 
+
+[2]: 1 distinct value 
+
+[2]: 1 distinct value
+
+[2,2]: 1 distinct value 
+
+The sum of the squares of the distinct counts in all subarrays is equal to 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> = 3.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>5</sup></code>

src/main/kotlin/g2901_3000/s2917_find_the_k_or_of_an_array/Solution.kt

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+package g2901_3000.s2917_find_the_k_or_of_an_array
+
+// #Easy #Array #Bit_Manipulation #2023_12_31_Time_191_ms_(76.47%)_Space_36.8_MB_(100.00%)
+
+class Solution {
+    fun findKOr(nums: IntArray, k: Int): Int {
+        val dp = IntArray(31)
+        for (num in nums) {
+            var i = 0
+            var localNum = num
+            while (localNum > 0) {
+                if ((localNum and 1) == 1) {
+                    dp[i] += 1
+                }
+                i += 1
+                localNum = localNum shr 1
+            }
+        }
+        var ans = 0
+        for (i in 0..30) {
+            if (dp[i] >= k) {
+                ans += (1 shl i)
+            }
+        }
+        return ans
+    }
+}

src/main/kotlin/g2901_3000/s2917_find_the_k_or_of_an_array/readme.md

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+2917\. Find the K-or of an Array
+
+Easy
+
+You are given a **0-indexed** integer array `nums`, and an integer `k`.
+
+The **K-or** of `nums` is a non-negative integer that satisfies the following:
+
+*   The <code>i<sup>th</sup></code> bit is set in the K-or **if and only if** there are at least `k` elements of nums in which bit `i` is set.
+
+Return _the **K-or** of_ `nums`.
+
+**Note** that a bit `i` is set in `x` if <code>(2<sup>i</sup> AND x) == 2<sup>i</sup></code>, where `AND` is the bitwise `AND` operator.
+
+**Example 1:**
+
+**Input:** nums = [7,12,9,8,9,15], k = 4
+
+**Output:** 9
+
+**Explanation:** 
+
+Bit 0 is set at nums[0], nums[2], nums[4], and nums[5]. 
+
+Bit 1 is set at nums[0], and nums[5]. 
+
+Bit 2 is set at nums[0], nums[1], and nums[5]. 
+
+Bit 3 is set at nums[1], nums[2], nums[3], nums[4], and nums[5]. 
+
+Only bits 0 and 3 are set in at least k elements of the array, and bits i >= 4 are not set in any of the array's elements. Hence, the answer is 2^0 + 2^3 = 9.
+
+**Example 2:**
+
+**Input:** nums = [2,12,1,11,4,5], k = 6
+
+**Output:** 0
+
+**Explanation:** Since k == 6 == nums.length, the 6-or of the array is equal to the bitwise AND of all its elements. Hence, the answer is 2 AND 12 AND 1 AND 11 AND 4 AND 5 = 0.
+
+**Example 3:**
+
+**Input:** nums = [10,8,5,9,11,6,8], k = 1
+
+**Output:** 15
+
+**Explanation:** Since k == 1, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 50`
+*   <code>0 <= nums[i] < 2<sup>31</sup></code>
+*   `1 <= k <= nums.length`

src/main/kotlin/g2901_3000/s2918_minimum_equal_sum_of_two_arrays_after_replacing_zeros/Solution.kt

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+package g2901_3000.s2918_minimum_equal_sum_of_two_arrays_after_replacing_zeros
+
+// #Medium #Array #Greedy #2023_12_31_Time_1410_ms_(7.69%)_Space_63.2_MB_(38.46%)
+
+class Solution {
+    fun minSum(nums1: IntArray, nums2: IntArray): Long {
+        val sum1 = nums1.fold(0L) { sum, element -> sum + element }
+        val zeroCount1 = nums1.count { it == 0 }
+
+        val sum2 = nums2.fold(0L) { sum, element -> sum + element }
+        val zeroCount2 = nums2.count { it == 0 }
+
+        if (
+            (zeroCount1 == 0 && sum1 < sum2 + zeroCount2) ||
+            (zeroCount2 == 0 && sum2 < sum1 + zeroCount1)
+        ) {
+            return -1
+        }
+        return Math.max(sum1 + zeroCount1, sum2 + zeroCount2)
+    }
+}

src/main/kotlin/g2901_3000/s2918_minimum_equal_sum_of_two_arrays_after_replacing_zeros/readme.md

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+2918\. Minimum Equal Sum of Two Arrays After Replacing Zeros
+
+Medium
+
+You are given two arrays `nums1` and `nums2` consisting of positive integers.
+
+You have to replace **all** the `0`'s in both arrays with **strictly** positive integers such that the sum of elements of both arrays becomes **equal**.
+
+Return _the **minimum** equal sum you can obtain, or_ `-1` _if it is impossible_.
+
+**Example 1:**
+
+**Input:** nums1 = [3,2,0,1,0], nums2 = [6,5,0]
+
+**Output:** 12
+
+**Explanation:** We can replace 0's in the following way: 
+- Replace the two 0's in nums1 with the values 2 and 4. The resulting array is nums1 = [3,2,2,1,4]. 
+- Replace the 0 in nums2 with the value 1. The resulting array is nums2 = [6,5,1]. Both arrays have an equal sum of 12. It can be shown that it is the minimum sum we can obtain.
+
+**Example 2:**
+
+**Input:** nums1 = [2,0,2,0], nums2 = [1,4]
+
+**Output:** -1
+
+**Explanation:** It is impossible to make the sum of both arrays equal.
+
+**Constraints:**
+
+*   <code>1 <= nums1.length, nums2.length <= 10<sup>5</sup></code>
+*   <code>0 <= nums1[i], nums2[i] <= 10<sup>6</sup></code>