Added tasks 3731-3734

Author: javadev

src/main/kotlin/g3701_3800/s3731_find_missing_elements/Solution.kt

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+package g3701_3800.s3731_find_missing_elements
+
+// #Easy #Weekly_Contest_474 #2025_11_03_Time_12_ms_(100.00%)_Space_48.81_MB_(11.11%)
+
+class Solution {
+    fun findMissingElements(nums: IntArray): MutableList<Int> {
+        val list: MutableList<Int> = ArrayList()
+        nums.sort()
+        for (i in 0..<nums.size - 1) {
+            if (nums[i + 1] - nums[i] > 1) {
+                for (j in nums[i] + 1..<nums[i + 1]) {
+                    list.add(j)
+                }
+            }
+        }
+        return list
+    }
+}

src/main/kotlin/g3701_3800/s3731_find_missing_elements/readme.md

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+3731\. Find Missing Elements
+
+Easy
+
+You are given an integer array `nums` consisting of **unique** integers.
+
+Originally, `nums` contained **every integer** within a certain range. However, some integers might have gone **missing** from the array.
+
+The **smallest** and **largest** integers of the original range are still present in `nums`.
+
+Return a **sorted** list of all the missing integers in this range. If no integers are missing, return an **empty** list.
+
+**Example 1:**
+
+**Input:** nums = [1,4,2,5]
+
+**Output:** [3]
+
+**Explanation:**
+
+The smallest integer is 1 and the largest is 5, so the full range should be `[1,2,3,4,5]`. Among these, only 3 is missing.
+
+**Example 2:**
+
+**Input:** nums = [7,8,6,9]
+
+**Output:** []
+
+**Explanation:**
+
+The smallest integer is 6 and the largest is 9, so the full range is `[6,7,8,9]`. All integers are already present, so no integer is missing.
+
+**Example 3:**
+
+**Input:** nums = [5,1]
+
+**Output:** [2,3,4]
+
+**Explanation:**
+
+The smallest integer is 1 and the largest is 5, so the full range should be `[1,2,3,4,5]`. The missing integers are 2, 3, and 4.
+
+**Constraints:**
+
+*   `2 <= nums.length <= 100`
+*   `1 <= nums[i] <= 100`

src/main/kotlin/g3701_3800/s3732_maximum_product_of_three_elements_after_one_replacement/Solution.kt

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+package g3701_3800.s3732_maximum_product_of_three_elements_after_one_replacement
+
+// #Medium #Weekly_Contest_474 #2025_11_03_Time_6_ms_(100.00%)_Space_75.20_MB_(76.92%)
+
+import kotlin.math.abs
+
+class Solution {
+    fun maxProduct(nums: IntArray): Long {
+        var a: Long = 0
+        var b: Long = 0
+        for (x in nums) {
+            val ax = abs(x).toLong()
+            if (ax >= a) {
+                b = a
+                a = ax
+            } else if (ax > b) {
+                b = ax
+            }
+        }
+        return 100000L * a * b
+    }
+}

src/main/kotlin/g3701_3800/s3732_maximum_product_of_three_elements_after_one_replacement/readme.md

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+3732\. Maximum Product of Three Elements After One Replacement
+
+Medium
+
+You are given an integer array `nums`.
+
+You **must** replace **exactly one** element in the array with **any** integer value in the range <code>[-10<sup>5</sup>, 10<sup>5</sup>]</code> (inclusive).
+
+After performing this single replacement, determine the **maximum possible product** of **any three** elements at **distinct indices** from the modified array.
+
+Return an integer denoting the **maximum product** achievable.
+
+**Example 1:**
+
+**Input:** nums = [-5,7,0]
+
+**Output:** 3500000
+
+**Explanation:**
+
+Replacing 0 with -10<sup>5</sup> gives the array <code>[-5, 7, -10<sup>5</sup>]</code>, which has a product <code>(-5) * 7 * (-10<sup>5</sup>) = 3500000</code>. The maximum product is 3500000.
+
+**Example 2:**
+
+**Input:** nums = [-4,-2,-1,-3]
+
+**Output:** 1200000
+
+**Explanation:**
+
+Two ways to achieve the maximum product include:
+
+*   `[-4, -2, -3]` → replace -2 with 10<sup>5</sup> → product = <code>(-4) * 10<sup>5</sup> * (-3) = 1200000</code>.
+*   `[-4, -1, -3]` → replace -1 with 10<sup>5</sup> → product = <code>(-4) * 10<sup>5</sup> * (-3) = 1200000</code>.
+
+The maximum product is 1200000.
+
+**Example 3:**
+
+**Input:** nums = [0,10,0]
+
+**Output:** 0
+
+**Explanation:**
+
+There is no way to replace an element with another integer and not have a 0 in the array. Hence, the product of all three elements will always be 0, and the maximum product is 0.
+
+**Constraints:**
+
+*   <code>3 <= nums.length <= 10<sup>5</sup></code>
+*   <code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code>

src/main/kotlin/g3701_3800/s3733_minimum_time_to_complete_all_deliveries/Solution.kt

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+package g3701_3800.s3733_minimum_time_to_complete_all_deliveries
+
+// #Medium #Weekly_Contest_474 #2025_11_03_Time_2_ms_(100.00%)_Space_42.94_MB_(100.00%)
+
+import kotlin.math.max
+
+class Solution {
+    private fun func(mid: Long, d: IntArray, r: IntArray): Boolean {
+        val lcm = r[0].toLong() * r[1] / gcd(r[0].toLong(), r[1].toLong())
+        val a1 = mid / r[0]
+        val a2 = mid / r[1]
+        val a3 = mid / lcm
+        val b = mid - a1 - a2 + a3
+        val o1 = a2 - a3
+        val o2 = a1 - a3
+        val d0 = max(d[0] - o1, 0)
+        val d1 = max(d[1] - o2, 0)
+        return b >= d0 + d1
+    }
+
+    private fun gcd(a: Long, b: Long): Long {
+        if (b == 0L) {
+            return a
+        }
+        return gcd(b, a % b)
+    }
+
+    fun minimumTime(d: IntArray, r: IntArray): Long {
+        var lo: Long = 0
+        var hi = 1e12.toLong()
+        var ans = Long.MAX_VALUE
+        while (lo <= hi) {
+            val mid = (lo + hi) / 2
+            if (func(mid, d, r)) {
+                ans = mid
+                hi = mid - 1
+            } else {
+                lo = mid + 1
+            }
+        }
+        return ans
+    }
+}

src/main/kotlin/g3701_3800/s3733_minimum_time_to_complete_all_deliveries/readme.md

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+3733\. Minimum Time to Complete All Deliveries
+
+Medium
+
+You are given two integer arrays of size 2: <code>d = [d<sub>1</sub>, d<sub>2</sub>]</code> and <code>r = [r<sub>1</sub>, r<sub>2</sub>]</code>.
+
+Two delivery drones are tasked with completing a specific number of deliveries. Drone `i` must complete <code>d<sub>i</sub></code> deliveries.
+
+Each delivery takes **exactly** one hour and **only one** drone can make a delivery at any given hour.
+
+Additionally, both drones require recharging at specific intervals during which they cannot make deliveries. Drone `i` must recharge every <code>r<sub>i</sub></code> hours (i.e. at hours that are multiples of <code>r<sub>i</sub></code>).
+
+Return an integer denoting the **minimum** total time (in hours) required to complete all deliveries.
+
+**Example 1:**
+
+**Input:** d = [3,1], r = [2,3]
+
+**Output:** 5
+
+**Explanation:**
+
+*   The first drone delivers at hours 1, 3, 5 (recharges at hours 2, 4).
+*   The second drone delivers at hour 2 (recharges at hour 3).
+
+**Example 2:**
+
+**Input:** d = [1,3], r = [2,2]
+
+**Output:** 7
+
+**Explanation:**
+
+*   The first drone delivers at hour 3 (recharges at hours 2, 4, 6).
+*   The second drone delivers at hours 1, 5, 7 (recharges at hours 2, 4, 6).
+
+**Example 3:**
+
+**Input:** d = [2,1], r = [3,4]
+
+**Output:** 3
+
+**Explanation:**
+
+*   The first drone delivers at hours 1, 2 (recharges at hour 3).
+*   The second drone delivers at hour 3.
+
+**Constraints:**
+
+*   <code>d = [d<sub>1</sub>, d<sub>2</sub>]</code>
+*   <code>1 <= d<sub>i</sub> <= 10<sup>9</sup></code>
+*   <code>r = [r<sub>1</sub>, r<sub>2</sub>]</code>
+*   <code>2 <= r<sub>i</sub> <= 3 * 10<sup>4</sup></code>

src/main/kotlin/g3701_3800/s3734_lexicographically_smallest_palindromic_permutation_greater_than_target/Solution.kt

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+package g3701_3800.s3734_lexicographically_smallest_palindromic_permutation_greater_than_target
+
+// #Hard #Weekly_Contest_474 #2025_11_03_Time_12_ms_(100.00%)_Space_47.24_MB_(100.00%)
+
+class Solution {
+    fun func(i: Int, target: String, ans: CharArray, l: Int, r: Int, freq: IntArray, end: Boolean): Boolean {
+        if (l > r) {
+            return String(ans).compareTo(target) > 0
+        }
+        if (l == r) {
+            var left = '#'
+            for (k in 0..25) {
+                if (freq[k] > 0) {
+                    left = (k + 'a'.code).toChar()
+                }
+            }
+            freq[left.code - 'a'.code]--
+            ans[l] = left
+            if (func(i + 1, target, ans, l + 1, r - 1, freq, end)) {
+                return true
+            }
+            freq[left.code - 'a'.code]++
+            ans[l] = '#'
+            return false
+        }
+        if (end) {
+            for (j in 0..25) {
+                if (freq[j] > 1) {
+                    freq[j] -= 2
+                    ans[r] = (j + 'a'.code).toChar()
+                    ans[l] = ans[r]
+                    if (func(i + 1, target, ans, l + 1, r - 1, freq, end)) {
+                        return true
+                    }
+                    ans[r] = '#'
+                    ans[l] = ans[r]
+                    freq[j] += 2
+                }
+            }
+            return false
+        }
+        val curr = target.get(i)
+        var next = '1'
+        for (k in target.get(i).code - 'a'.code + 1..25) {
+            if (freq[k] > 1) {
+                next = (k + 'a'.code).toChar()
+                break
+            }
+        }
+        if (freq[curr.code - 'a'.code] > 1) {
+            ans[r] = curr
+            ans[l] = ans[r]
+            freq[curr.code - 'a'.code] -= 2
+            if (func(i + 1, target, ans, l + 1, r - 1, freq, end)) {
+                return true
+            }
+            freq[curr.code - 'a'.code] += 2
+        }
+        if (next != '1') {
+            ans[r] = next
+            ans[l] = ans[r]
+            freq[next.code - 'a'.code] -= 2
+            if (func(i + 1, target, ans, l + 1, r - 1, freq, true)) {
+                return true
+            }
+            freq[next.code - 'a'.code] += 2
+        }
+        ans[r] = '#'
+        ans[l] = ans[r]
+        return false
+    }
+
+    fun lexPalindromicPermutation(s: String, target: String): String {
+        val freq = IntArray(26)
+        for (i in 0..<s.length) {
+            freq[s.get(i).code - 'a'.code]++
+        }
+        var oddc = 0
+        for (i in 0..25) {
+            if (freq[i] % 2 == 1) {
+                oddc++
+            }
+        }
+        if (oddc > 1) {
+            return ""
+        }
+        val ans = CharArray(s.length)
+        ans.fill('#')
+        if (func(0, target, ans, 0, s.length - 1, freq, false)) {
+            return String(ans)
+        }
+        return ""
+    }
+}

src/main/kotlin/g3701_3800/s3734_lexicographically_smallest_palindromic_permutation_greater_than_target/readme.md

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+3734\. Lexicographically Smallest Palindromic Permutation Greater Than Target
+
+Hard
+
+You are given two strings `s` and `target`, each of length `n`, consisting of lowercase English letters.
+
+Return the **lexicographically smallest string** that is **both** a **palindromic permutation** of `s` and **strictly** greater than `target`. If no such permutation exists, return an empty string.
+
+**Example 1:**
+
+**Input:** s = "baba", target = "abba"
+
+**Output:** "baab"
+
+**Explanation:**
+
+*   The palindromic permutations of `s` (in lexicographical order) are `"abba"` and `"baab"`.
+*   The lexicographically smallest permutation that is strictly greater than `target` is `"baab"`.
+
+**Example 2:**
+
+**Input:** s = "baba", target = "bbaa"
+
+**Output:** ""
+
+**Explanation:**
+
+*   The palindromic permutations of `s` (in lexicographical order) are `"abba"` and `"baab"`.
+*   None of them is lexicographically strictly greater than `target`. Therefore, the answer is `""`.
+
+**Example 3:**
+
+**Input:** s = "abc", target = "abb"
+
+**Output:** ""
+
+**Explanation:**
+
+`s` has no palindromic permutations. Therefore, the answer is `""`.
+
+**Example 4:**
+
+**Input:** s = "aac", target = "abb"
+
+**Output:** "aca"
+
+**Explanation:**
+
+*   The only palindromic permutation of `s` is `"aca"`.
+*   `"aca"` is strictly greater than `target`. Therefore, the answer is `"aca"`.
+
+**Constraints:**
+
+*   `1 <= n == s.length == target.length <= 300`
+*   `s` and `target` consist of only lowercase English letters.