Added tasks 3536-3539

Author: javadev

src/main/kotlin/g3501_3600/s3536_maximum_product_of_two_digits/Solution.kt

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+package g3501_3600.s3536_maximum_product_of_two_digits
+
+// #Easy #2025_05_04_Time_1_ms_(100.00%)_Space_40.93_MB_(100.00%)
+
+class Solution {
+    fun maxProduct(n: Int): Int {
+        var n = n
+        var m1 = n % 10
+        n /= 10
+        var m2 = n % 10
+        n /= 10
+        while (n > 0) {
+            val a = n % 10
+            if (a > m1) {
+                if (m1 > m2) {
+                    m2 = m1
+                }
+                m1 = a
+            } else {
+                if (a > m2) {
+                    m2 = a
+                }
+            }
+            n /= 10
+        }
+        return m1 * m2
+    }
+}

src/main/kotlin/g3501_3600/s3536_maximum_product_of_two_digits/readme.md

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+3536\. Maximum Product of Two Digits
+
+Easy
+
+You are given a positive integer `n`.
+
+Return the **maximum** product of any two digits in `n`.
+
+**Note:** You may use the **same** digit twice if it appears more than once in `n`.
+
+**Example 1:**
+
+**Input:** n = 31
+
+**Output:** 3
+
+**Explanation:**
+
+*   The digits of `n` are `[3, 1]`.
+*   The possible products of any two digits are: `3 * 1 = 3`.
+*   The maximum product is 3.
+
+**Example 2:**
+
+**Input:** n = 22
+
+**Output:** 4
+
+**Explanation:**
+
+*   The digits of `n` are `[2, 2]`.
+*   The possible products of any two digits are: `2 * 2 = 4`.
+*   The maximum product is 4.
+
+**Example 3:**
+
+**Input:** n = 124
+
+**Output:** 8
+
+**Explanation:**
+
+*   The digits of `n` are `[1, 2, 4]`.
+*   The possible products of any two digits are: `1 * 2 = 2`, `1 * 4 = 4`, `2 * 4 = 8`.
+*   The maximum product is 8.
+
+**Constraints:**
+
+*   <code>10 <= n <= 10<sup>9</sup></code>

src/main/kotlin/g3501_3600/s3537_fill_a_special_grid/Solution.kt

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+package g3501_3600.s3537_fill_a_special_grid
+
+// #Medium #2025_05_04_Time_2_ms_(100.00%)_Space_88.71_MB_(61.54%)
+
+import kotlin.math.pow
+
+class Solution {
+    fun specialGrid(n: Int): Array<IntArray> {
+        if (n == 0) {
+            return arrayOf<IntArray>(intArrayOf(0))
+        }
+        val len = 2.0.pow(n.toDouble()).toInt()
+        val ans = Array<IntArray>(len) { IntArray(len) }
+        val num = intArrayOf(2.0.pow(2.0 * n).toInt() - 1)
+        backtrack(ans, len, len, 0, 0, num)
+        return ans
+    }
+
+    private fun backtrack(ans: Array<IntArray>, m: Int, n: Int, x: Int, y: Int, num: IntArray) {
+        if (m == 2 && n == 2) {
+            ans[x][y] = num[0]
+            ans[x + 1][y] = num[0] - 1
+            ans[x + 1][y + 1] = num[0] - 2
+            ans[x][y + 1] = num[0] - 3
+            num[0] -= 4
+            return
+        }
+        backtrack(ans, m / 2, n / 2, x, y, num)
+        backtrack(ans, m / 2, n / 2, x + m / 2, y, num)
+        backtrack(ans, m / 2, n / 2, x + m / 2, y + n / 2, num)
+        backtrack(ans, m / 2, n / 2, x, y + n / 2, num)
+    }
+}

src/main/kotlin/g3501_3600/s3537_fill_a_special_grid/readme.md

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+3537\. Fill a Special Grid
+
+Medium
+
+You are given a non-negative integer `n` representing a <code>2<sup>n</sup> x 2<sup>n</sup></code> grid. You must fill the grid with integers from 0 to <code>2<sup>2n</sup> - 1</code> to make it **special**. A grid is **special** if it satisfies **all** the following conditions:
+
+*   All numbers in the top-right quadrant are smaller than those in the bottom-right quadrant.
+*   All numbers in the bottom-right quadrant are smaller than those in the bottom-left quadrant.
+*   All numbers in the bottom-left quadrant are smaller than those in the top-left quadrant.
+*   Each of its quadrants is also a special grid.
+
+Return the **special** <code>2<sup>n</sup> x 2<sup>n</sup></code> grid.
+
+**Note**: Any 1x1 grid is special.
+
+**Example 1:**
+
+**Input:** n = 0
+
+**Output:** [[0]]
+
+**Explanation:**
+
+The only number that can be placed is 0, and there is only one possible position in the grid.
+
+**Example 2:**
+
+**Input:** n = 1
+
+**Output:** [[3,0],[2,1]]
+
+**Explanation:**
+
+The numbers in each quadrant are:
+
+*   Top-right: 0
+*   Bottom-right: 1
+*   Bottom-left: 2
+*   Top-left: 3
+
+Since `0 < 1 < 2 < 3`, this satisfies the given constraints.
+
+**Example 3:**
+
+**Input:** n = 2
+
+**Output:** [[15,12,3,0],[14,13,2,1],[11,8,7,4],[10,9,6,5]]
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2025/03/05/4123example3p1drawio.png)
+
+The numbers in each quadrant are:
+
+*   Top-right: 3, 0, 2, 1
+*   Bottom-right: 7, 4, 6, 5
+*   Bottom-left: 11, 8, 10, 9
+*   Top-left: 15, 12, 14, 13
+*   `max(3, 0, 2, 1) < min(7, 4, 6, 5)`
+*   `max(7, 4, 6, 5) < min(11, 8, 10, 9)`
+*   `max(11, 8, 10, 9) < min(15, 12, 14, 13)`
+
+This satisfies the first three requirements. Additionally, each quadrant is also a special grid. Thus, this is a special grid.
+
+**Constraints:**
+
+*   `0 <= n <= 10`

src/main/kotlin/g3501_3600/s3538_merge_operations_for_minimum_travel_time/Solution.kt

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+package g3501_3600.s3538_merge_operations_for_minimum_travel_time
+
+// #Hard #2025_05_04_Time_10_ms_(100.00%)_Space_46.96_MB_(100.00%)
+
+import kotlin.math.min
+
+@Suppress("unused")
+class Solution {
+    fun minTravelTime(l: Int, n: Int, k: Int, position: IntArray, time: IntArray): Int {
+        val dp = Array<Array<IntArray>>(n) { Array<IntArray>(k + 1) { IntArray(k + 1) } }
+        for (i in 0..<n) {
+            for (j in 0..k) {
+                for (m in 0..k) {
+                    dp[i][j][m] = Int.Companion.MAX_VALUE
+                }
+            }
+        }
+        dp[0][0][0] = 0
+        for (i in 0..<n - 1) {
+            var currTime = 0
+            var curr = 0
+            while (curr <= k && curr <= i) {
+                currTime += time[i - curr]
+                for (used in 0..k) {
+                    if (dp[i][curr][used] == Int.Companion.MAX_VALUE) {
+                        continue
+                    }
+                    var next = 0
+                    while (next <= k - used && next <= n - i - 2) {
+                        val nextI = i + next + 1
+                        dp[nextI][next][next + used] = min(
+                            dp[nextI][next][next + used],
+                            dp[i][curr][used] +
+                                (position[nextI] - position[i]) * currTime,
+                        )
+                        next++
+                    }
+                }
+                curr++
+            }
+        }
+        var ans = Int.Companion.MAX_VALUE
+        for (curr in 0..k) {
+            ans = min(ans, dp[n - 1][curr][k])
+        }
+        return ans
+    }
+}

src/main/kotlin/g3501_3600/s3538_merge_operations_for_minimum_travel_time/readme.md

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+3538\. Merge Operations for Minimum Travel Time
+
+Hard
+
+You are given a straight road of length `l` km, an integer `n`, an integer `k`**,** and **two** integer arrays, `position` and `time`, each of length `n`.
+
+The array `position` lists the positions (in km) of signs in **strictly** increasing order (with `position[0] = 0` and `position[n - 1] = l`).
+
+Each `time[i]` represents the time (in minutes) required to travel 1 km between `position[i]` and `position[i + 1]`.
+
+You **must** perform **exactly** `k` merge operations. In one merge, you can choose any **two** adjacent signs at indices `i` and `i + 1` (with `i > 0` and `i + 1 < n`) and:
+
+*   Update the sign at index `i + 1` so that its time becomes `time[i] + time[i + 1]`.
+*   Remove the sign at index `i`.
+
+Return the **minimum** **total** **travel time** (in minutes) to travel from 0 to `l` after **exactly** `k` merges.
+
+**Example 1:**
+
+**Input:** l = 10, n = 4, k = 1, position = [0,3,8,10], time = [5,8,3,6]
+
+**Output:** 62
+
+**Explanation:**
+
+*   Merge the signs at indices 1 and 2. Remove the sign at index 1, and change the time at index 2 to `8 + 3 = 11`.
+    
+*   After the merge:
+    *   `position` array: `[0, 8, 10]`
+    *   `time` array: `[5, 11, 6]`
+    
+| Segment   | Distance (km) | Time per km (min) | Segment Travel Time (min)  |
+|-----------|---------------|-------------------|----------------------------|
+| 0 → 8     | 8             | 5                 | 8 × 5 = 40                 |
+| 8 → 10    | 2             | 11                | 2 × 11 = 22                |
+
+    
+*   Total Travel Time: `40 + 22 = 62`, which is the minimum possible time after exactly 1 merge.
+
+**Example 2:**
+
+**Input:** l = 5, n = 5, k = 1, position = [0,1,2,3,5], time = [8,3,9,3,3]
+
+**Output:** 34
+
+**Explanation:**
+
+*   Merge the signs at indices 1 and 2. Remove the sign at index 1, and change the time at index 2 to `3 + 9 = 12`.
+*   After the merge:
+    *   `position` array: `[0, 2, 3, 5]`
+    *   `time` array: `[8, 12, 3, 3]`
+    
+| Segment   | Distance (km) | Time per km (min) | Segment Travel Time (min)  |
+|-----------|---------------|-------------------|----------------------------|
+| 0 → 2     | 2             | 8                 | 2 × 8 = 16                 |
+| 2 → 3     | 1             | 12                | 1 × 12 = 12                |
+| 3 → 5     | 2             | 3                 | 2 × 3 = 6                  |
+    
+*   Total Travel Time: `16 + 12 + 6 = 34`**,** which is the minimum possible time after exactly 1 merge.
+
+**Constraints:**
+
+*   <code>1 <= l <= 10<sup>5</sup></code>
+*   `2 <= n <= min(l + 1, 50)`
+*   `0 <= k <= min(n - 2, 10)`
+*   `position.length == n`
+*   `position[0] = 0` and `position[n - 1] = l`
+*   `position` is sorted in strictly increasing order.
+*   `time.length == n`
+*   `1 <= time[i] <= 100`
+*   `1 <= sum(time) <= 100`

src/main/kotlin/g3501_3600/s3539_find_sum_of_array_product_of_magical_sequences/Solution.kt

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+package g3501_3600.s3539_find_sum_of_array_product_of_magical_sequences
+
+// #Hard #2025_05_04_Time_58_ms_(_%)_Space_49.73_MB_(_%)
+
+class Solution {
+    fun magicalSum(m: Int, k: Int, nums: IntArray): Int {
+        val n = nums.size
+        val pow = Array<LongArray>(n) { LongArray(m + 1) }
+        for (j in 0..<n) {
+            pow[j][0] = 1L
+            for (c in 1..m) {
+                pow[j][c] = pow[j][c - 1] * nums[j] % MOD
+            }
+        }
+        var dp = Array<Array<LongArray>>(m + 1) { Array<LongArray>(k + 1) { LongArray(m + 1) } }
+        var next = Array<Array<LongArray>>(m + 1) { Array<LongArray>(k + 1) { LongArray(m + 1) } }
+        dp[0][0][0] = 1L
+        for (i in 0..<n) {
+            for (t in 0..m) {
+                for (o in 0..k) {
+                    next[t][o].fill(0L)
+                }
+            }
+            for (t in 0..m) {
+                for (o in 0..k) {
+                    for (c in 0..m) {
+                        if (dp[t][o][c] == 0L) {
+                            continue
+                        }
+                        for (cc in 0..m - t) {
+                            val total = c + cc
+                            if (o + (total and 1) > k) {
+                                continue
+                            }
+                            next[t + cc][o + (total and 1)][total ushr 1] =
+                                (
+                                    (
+                                        next[t + cc][o + (total and 1)][total ushr 1] +
+                                            dp[t][o][c] *
+                                            C[m - t][cc] %
+                                            MOD
+                                            * pow[i][cc] %
+                                            MOD
+                                        ) %
+                                        MOD
+                                    )
+                        }
+                    }
+                }
+            }
+            val tmp = dp
+            dp = next
+            next = tmp
+        }
+        var res: Long = 0
+        for (o in 0..k) {
+            for (c in 0..m) {
+                if (o + P[c] == k) {
+                    res = (res + dp[m][o][c]) % MOD
+                }
+            }
+        }
+        return res.toInt()
+    }
+
+    companion object {
+        private const val MOD = 1000000007
+        private val C: Array<IntArray> = precomputeBinom(31)
+        private val P: IntArray = precomputePop(31)
+
+        private fun precomputeBinom(max: Int): Array<IntArray> {
+            val res = Array<IntArray>(max) { IntArray(max) }
+            for (i in 0..<max) {
+                res[i][i] = 1
+                res[i][0] = res[i][i]
+                for (j in 1..<i) {
+                    res[i][j] = (res[i - 1][j - 1] + res[i - 1][j]) % MOD
+                }
+            }
+            return res
+        }
+
+        private fun precomputePop(max: Int): IntArray {
+            val res = IntArray(max)
+            for (i in 1..<max) {
+                res[i] = res[i shr 1] + (i and 1)
+            }
+            return res
+        }
+    }
+}