Added tasks 3487-3490

Author: javadev

src/main/kotlin/g3401_3500/s3487_maximum_unique_subarray_sum_after_deletion/Solution.kt

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+package g3401_3500.s3487_maximum_unique_subarray_sum_after_deletion
+
+// #Easy #2025_03_16_Time_4_ms_(100.00%)_Space_43.27_MB_(100.00%)
+
+class Solution {
+    fun maxSum(nums: IntArray): Int {
+        var sum = 0
+        val st = mutableSetOf<Int>()
+        var mxNeg = Int.MIN_VALUE
+        for (num in nums) {
+            if (num > 0) {
+                st.add(num)
+            } else {
+                mxNeg = maxOf(mxNeg, num)
+            }
+        }
+        for (value in st) {
+            sum += value
+        }
+        return if (st.isNotEmpty()) sum else mxNeg
+    }
+}

src/main/kotlin/g3401_3500/s3487_maximum_unique_subarray_sum_after_deletion/readme.md

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+3487\. Maximum Unique Subarray Sum After Deletion
+
+Easy
+
+You are given an integer array `nums`.
+
+You are allowed to delete any number of elements from `nums` without making it **empty**. After performing the deletions, select a non-empty subarrays of `nums` such that:
+
+1.  All elements in the subarray are **unique**.
+2.  The sum of the elements in the subarray is **maximized**.
+
+Return the **maximum sum** of such a subarray.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,4,5]
+
+**Output:** 15
+
+**Explanation:**
+
+Select the entire array without deleting any element to obtain the maximum sum.
+
+**Example 2:**
+
+**Input:** nums = [1,1,0,1,1]
+
+**Output:** 1
+
+**Explanation:**
+
+Delete the element `nums[0] == 1`, `nums[1] == 1`, `nums[2] == 0`, and `nums[3] == 1`. Select the entire array `[1]` to obtain the maximum sum.
+
+**Example 3:**
+
+**Input:** nums = [1,2,-1,-2,1,0,-1]
+
+**Output:** 3
+
+**Explanation:**
+
+Delete the elements `nums[2] == -1` and `nums[3] == -2`, and select the subarray `[2, 1]` from `[1, 2, 1, 0, -1]` to obtain the maximum sum.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `-100 <= nums[i] <= 100`

src/main/kotlin/g3401_3500/s3488_closest_equal_element_queries/Solution.kt

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+package g3401_3500.s3488_closest_equal_element_queries
+
+// #Medium #2025_03_16_Time_93_ms_(100.00%)_Space_99.42_MB_(100.00%)
+
+import kotlin.math.abs
+import kotlin.math.min
+
+class Solution {
+    fun solveQueries(nums: IntArray, queries: IntArray): List<Int> {
+        val sz = nums.size
+        val indices: MutableMap<Int, MutableList<Int>> = HashMap<Int, MutableList<Int>>()
+        for (i in 0..<sz) {
+            indices.computeIfAbsent(nums[i]) { _: Int -> ArrayList<Int>() }.add(i)
+        }
+        for (arr in indices.values) {
+            val m = arr.size
+            if (m == 1) {
+                nums[arr[0]] = -1
+                continue
+            }
+            for (i in 0..<m) {
+                val j: Int = arr[i]
+                val f: Int = arr[(i + 1) % m]
+                val b: Int = arr[(i - 1 + m) % m]
+                val forward = min(((sz - j - 1) + f + 1), abs((j - f)))
+                val backward = min(abs((b - j)), (j + (sz - b)))
+                nums[j] = min(backward, forward)
+            }
+        }
+        val res: MutableList<Int> = ArrayList<Int>()
+        for (q in queries) {
+            res.add(nums[q])
+        }
+        return res
+    }
+}

src/main/kotlin/g3401_3500/s3488_closest_equal_element_queries/readme.md

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+3488\. Closest Equal Element Queries
+
+Medium
+
+You are given a **circular** array `nums` and an array `queries`.
+
+For each query `i`, you have to find the following:
+
+*   The **minimum** distance between the element at index `queries[i]` and **any** other index `j` in the **circular** array, where `nums[j] == nums[queries[i]]`. If no such index exists, the answer for that query should be -1.
+
+Return an array `answer` of the **same** size as `queries`, where `answer[i]` represents the result for query `i`.
+
+**Example 1:**
+
+**Input:** nums = [1,3,1,4,1,3,2], queries = [0,3,5]
+
+**Output:** [2,-1,3]
+
+**Explanation:**
+
+*   Query 0: The element at `queries[0] = 0` is `nums[0] = 1`. The nearest index with the same value is 2, and the distance between them is 2.
+*   Query 1: The element at `queries[1] = 3` is `nums[3] = 4`. No other index contains 4, so the result is -1.
+*   Query 2: The element at `queries[2] = 5` is `nums[5] = 3`. The nearest index with the same value is 1, and the distance between them is 3 (following the circular path: `5 -> 6 -> 0 -> 1`).
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,4], queries = [0,1,2,3]
+
+**Output:** [-1,-1,-1,-1]
+
+**Explanation:**
+
+Each value in `nums` is unique, so no index shares the same value as the queried element. This results in -1 for all queries.
+
+**Constraints:**
+
+*   <code>1 <= queries.length <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>6</sup></code>
+*   `0 <= queries[i] < nums.length`

src/main/kotlin/g3401_3500/s3489_zero_array_transformation_iv/Solution.kt

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+package g3401_3500.s3489_zero_array_transformation_iv
+
+// #Medium #2025_03_16_Time_104_ms_(100.00%)_Space_73.10_MB_(100.00%)
+
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    private fun solve(q: Array<IntArray>, i: Int, target: Int, k: Int, dp: Array<IntArray>): Int {
+        // we found  a valid sum equal to target so return current index of query.
+        if (target == 0) {
+            return k
+        }
+        // return a larger number to invalidate this flow
+        if (k >= q.size || target < 0) {
+            return q.size + 1
+        }
+        if (dp[target][k] != -1) {
+            return dp[target][k]
+        }
+        // skip current query val
+        var res = solve(q, i, target, k + 1, dp)
+        // pick the val if its range is in the range of target index
+        if (q[k][0] <= i && i <= q[k][1]) {
+            res = min(res, solve(q, i, target - q[k][2], k + 1, dp))
+        }
+        dp[target][k] = res
+        return res
+    }
+
+    fun minZeroArray(nums: IntArray, queries: Array<IntArray>): Int {
+        var ans = -1
+        for (i in nums.indices) {
+            val dp = Array<IntArray>(nums[i] + 1) { IntArray(queries.size) }
+            dp.forEach { row: IntArray -> row.fill(-1) }
+            ans = max(ans, solve(queries, i, nums[i], 0, dp))
+        }
+        return if (ans > queries.size) -1 else ans
+    }
+}

src/main/kotlin/g3401_3500/s3489_zero_array_transformation_iv/readme.md

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+3489\. Zero Array Transformation IV
+
+Medium
+
+You are given an integer array `nums` of length `n` and a 2D array `queries`, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>, val<sub>i</sub>]</code>.
+
+Each `queries[i]` represents the following action on `nums`:
+
+*   Select a **subset** of indices in the range <code>[l<sub>i</sub>, r<sub>i</sub>]</code> from `nums`.
+*   Decrement the value at each selected index by **exactly** <code>val<sub>i</sub></code>.
+
+A **Zero Array** is an array with all its elements equal to 0.
+
+Return the **minimum** possible **non-negative** value of `k`, such that after processing the first `k` queries in **sequence**, `nums` becomes a **Zero Array**. If no such `k` exists, return -1.
+
+**Example 1:**
+
+**Input:** nums = [2,0,2], queries = [[0,2,1],[0,2,1],[1,1,3]]
+
+**Output:** 2
+
+**Explanation:**
+
+*   **For query 0 (l = 0, r = 2, val = 1):**
+    *   Decrement the values at indices `[0, 2]` by 1.
+    *   The array will become `[1, 0, 1]`.
+*   **For query 1 (l = 0, r = 2, val = 1):**
+    *   Decrement the values at indices `[0, 2]` by 1.
+    *   The array will become `[0, 0, 0]`, which is a Zero Array. Therefore, the minimum value of `k` is 2.
+
+**Example 2:**
+
+**Input:** nums = [4,3,2,1], queries = [[1,3,2],[0,2,1]]
+
+**Output:** \-1
+
+**Explanation:**
+
+It is impossible to make nums a Zero Array even after all the queries.
+
+**Example 3:**
+
+**Input:** nums = [1,2,3,2,1], queries = [[0,1,1],[1,2,1],[2,3,2],[3,4,1],[4,4,1]]
+
+**Output:** 4
+
+**Explanation:**
+
+*   **For query 0 (l = 0, r = 1, val = 1):**
+    *   Decrement the values at indices `[0, 1]` by `1`.
+    *   The array will become `[0, 1, 3, 2, 1]`.
+*   **For query 1 (l = 1, r = 2, val = 1):**
+    *   Decrement the values at indices `[1, 2]` by 1.
+    *   The array will become `[0, 0, 2, 2, 1]`.
+*   **For query 2 (l = 2, r = 3, val = 2):**
+    *   Decrement the values at indices `[2, 3]` by 2.
+    *   The array will become `[0, 0, 0, 0, 1]`.
+*   **For query 3 (l = 3, r = 4, val = 1):**
+    *   Decrement the value at index 4 by 1.
+    *   The array will become `[0, 0, 0, 0, 0]`. Therefore, the minimum value of `k` is 4.
+
+**Example 4:**
+
+**Input:** nums = [1,2,3,2,6], queries = [[0,1,1],[0,2,1],[1,4,2],[4,4,4],[3,4,1],[4,4,5]]
+
+**Output:** 4
+
+**Constraints:**
+
+*   `1 <= nums.length <= 10`
+*   `0 <= nums[i] <= 1000`
+*   `1 <= queries.length <= 1000`
+*   <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>, val<sub>i</sub>]</code>
+*   <code>0 <= l<sub>i</sub> <= r<sub>i</sub> < nums.length</code>
+*   <code>1 <= val<sub>i</sub> <= 10</code>

src/main/kotlin/g3401_3500/s3490_count_beautiful_numbers/Solution.kt

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+package g3401_3500.s3490_count_beautiful_numbers
+
+// #Hard #2025_03_16_Time_246_ms_(100.00%)_Space_61.00_MB_(100.00%)
+
+class Solution {
+    fun beautifulNumbers(l: Int, r: Int): Int {
+        return countBeautiful(r) - countBeautiful(l - 1)
+    }
+
+    private fun countBeautiful(x: Int): Int {
+        val digits = getCharArray(x)
+        val dp = HashMap<String?, Int?>()
+        return solve(0, 1, 0, 1, digits, dp)
+    }
+
+    private fun getCharArray(x: Int): CharArray {
+        val str = x.toString()
+        return str.toCharArray()
+    }
+
+    private fun solve(
+        i: Int,
+        tight: Int,
+        sum: Int,
+        prod: Int,
+        digits: CharArray,
+        dp: HashMap<String?, Int?>,
+    ): Int {
+        if (i == digits.size) {
+            return if (sum > 0 && prod % sum == 0) {
+                1
+            } else {
+                0
+            }
+        }
+        val str = "$i - $tight - $sum - $prod"
+        if (dp.containsKey(str)) {
+            return dp.get(str)!!
+        }
+        val limit: Int = if (tight == 1) {
+            digits[i].code - '0'.code
+        } else {
+            9
+        }
+        var count = 0
+        var j = 0
+        while (j <= limit) {
+            var newTight = 0
+            if (tight == 1 && j == limit) {
+                newTight = 1
+            }
+            val newSum = sum + j
+            val newProd: Int = if (j == 0 && sum == 0) {
+                1
+            } else {
+                prod * j
+            }
+            count += solve(i + 1, newTight, newSum, newProd, digits, dp)
+            j++
+        }
+        dp.put(str, count)
+        return count
+    }
+}

src/main/kotlin/g3401_3500/s3490_count_beautiful_numbers/readme.md

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+3490\. Count Beautiful Numbers
+
+Hard
+
+You are given two positive integers, `l` and `r`. A positive integer is called **beautiful** if the product of its digits is divisible by the sum of its digits.
+
+Return the count of **beautiful** numbers between `l` and `r`, inclusive.
+
+**Example 1:**
+
+**Input:** l = 10, r = 20
+
+**Output:** 2
+
+**Explanation:**
+
+The beautiful numbers in the range are 10 and 20.
+
+**Example 2:**
+
+**Input:** l = 1, r = 15
+
+**Output:** 10
+
+**Explanation:**
+
+The beautiful numbers in the range are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.
+
+**Constraints:**
+
+*   <code>1 <= l <= r < 10<sup>9</sup></code>

src/test/kotlin/g3401_3500/s3487_maximum_unique_subarray_sum_after_deletion/SolutionTest.kt

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+package g3401_3500.s3487_maximum_unique_subarray_sum_after_deletion
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun maxSum() {
+        assertThat<Int>(Solution().maxSum(intArrayOf(1, 2, 3, 4, 5)), equalTo<Int>(15))
+    }
+
+    @Test
+    fun maxSum2() {
+        assertThat<Int>(Solution().maxSum(intArrayOf(1, 1, 0, 1, 1)), equalTo<Int>(1))
+    }
+
+    @Test
+    fun maxSum3() {
+        assertThat<Int>(
+            Solution().maxSum(intArrayOf(1, 2, -1, -2, 1, 0, -1)),
+            equalTo<Int>(3),
+        )
+    }
+
+    @Test
+    fun maxSum4() {
+        assertThat<Int>(Solution().maxSum(intArrayOf(-100)), equalTo<Int>(-100))
+    }
+}