Added tasks 3731-3734

src/main/java/g3701_3800/s3731_find_missing_elements/Solution.java

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+package g3701_3800.s3731_find_missing_elements;
+
+// #Easy #Array #Hash_Table #Sorting #Weekly_Contest_474
+// #2025_11_05_Time_2_ms_(94.89%)_Space_46.54_MB_(95.16%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    public List<Integer> findMissingElements(int[] nums) {
+        int maxi = 0;
+        int mini = 101;
+        List<Integer> list = new ArrayList<>();
+        boolean[] array = new boolean[101];
+        for (int num : nums) {
+            array[num] = true;
+            if (maxi < num) {
+                maxi = num;
+            }
+            if (mini > num) {
+                mini = num;
+            }
+        }
+        for (int index = mini + 1; index < maxi; index++) {
+            if (array[index]) {
+                continue;
+            }
+            list.add(index);
+        }
+        return list;
+    }
+}

src/main/java/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/java/g3701_3800/s3732_maximum_product_of_three_elements_after_one_replacement/Solution.java

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+package g3701_3800.s3732_maximum_product_of_three_elements_after_one_replacement;
+
+// #Medium #Array #Math #Sorting #Greedy #Weekly_Contest_474
+// #2025_11_05_Time_4_ms_(95.32%)_Space_97.32_MB_(28.84%)
+
+public class Solution {
+    public long maxProduct(int[] nums) {
+        long a = 0;
+        long b = 0;
+        for (int x : nums) {
+            long ax = Math.abs(x);
+            if (ax >= a) {
+                b = a;
+                a = ax;
+            } else if (ax > b) {
+                b = ax;
+            }
+        }
+        return 100000L * a * b;
+    }
+}

src/main/java/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/java/g3701_3800/s3733_minimum_time_to_complete_all_deliveries/Solution.java

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+package g3701_3800.s3733_minimum_time_to_complete_all_deliveries;
+
+// #Medium #Math #Binary_Search #Weekly_Contest_474
+// #2025_11_05_Time_1_ms_(100.00%)_Space_44.16_MB_(55.61%)
+
+public class Solution {
+    private boolean pos(long k, long n1, long n2, int p1, int p2, long lVal) {
+        long kP1 = k / p1;
+        long kP2 = k / p2;
+        long kL = k / lVal;
+        long s1 = kP2 - kL;
+        long s2 = kP1 - kL;
+        long sB = k - kP1 - kP2 + kL;
+        long w1 = Math.max(0L, n1 - s1);
+        long w2 = Math.max(0L, n2 - s2);
+        return (w1 + w2) <= sB;
+    }
+
+    private long findGcd(long a, long b) {
+        if (b == 0) {
+            return a;
+        }
+        return findGcd(b, a % b);
+    }
+
+    public long minimumTime(int[] d, int[] r) {
+        long n1 = d[0];
+        long n2 = d[1];
+        int p1 = r[0];
+        int p2 = r[1];
+        long g = findGcd(p1, p2);
+        long l = (long) p1 * p2 / g;
+        long lo = n1 + n2;
+        long hi = (n1 + n2) * Math.max(p1, p2);
+        long res = hi;
+        while (lo <= hi) {
+            long mid = lo + (hi - lo) / 2;
+            if (pos(mid, n1, n2, p1, p2, l)) {
+                res = mid;
+                hi = mid - 1;
+            } else {
+                lo = mid + 1;
+            }
+        }
+        return res;
+    }
+}

src/main/java/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/java/g3701_3800/s3734_lexicographically_smallest_palindromic_permutation_greater_than_target/Solution.java

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+package g3701_3800.s3734_lexicographically_smallest_palindromic_permutation_greater_than_target;
+
+// #Hard #String #Two_Pointers #Enumeration #Weekly_Contest_474
+// #2025_11_05_Time_2_ms_(100.00%)_Space_46.34_MB_(84.73%)
+
+import java.util.Arrays;
+
+public class Solution {
+    boolean func(int i, String target, char[] ans, int l, int r, int[] freq, boolean end) {
+        if (l > r) {
+            return new String(ans).compareTo(target) > 0;
+        }
+        if (l == r) {
+            char left = '#';
+            for (int k = 0; k < 26; k++) {
+                if (freq[k] > 0) {
+                    left = (char) (k + 'a');
+                }
+            }
+            freq[left - 'a']--;
+            ans[l] = left;
+            if (func(i + 1, target, ans, l + 1, r - 1, freq, end)) {
+                return true;
+            }
+            freq[left - 'a']++;
+            ans[l] = '#';
+            return false;
+        }
+        if (end) {
+            for (int j = 0; j < 26; j++) {
+                if (freq[j] > 1) {
+                    freq[j] -= 2;
+                    ans[l] = ans[r] = (char) (j + 'a');
+                    if (func(i + 1, target, ans, l + 1, r - 1, freq, end)) {
+                        return true;
+                    }
+                    ans[l] = ans[r] = '#';
+                    freq[j] += 2;
+                }
+            }
+            return false;
+        }
+        char curr = target.charAt(i);
+        char next = '1';
+        for (int k = target.charAt(i) - 'a' + 1; k < 26; k++) {
+            if (freq[k] > 1) {
+                next = (char) (k + 'a');
+                break;
+            }
+        }
+        if (freq[curr - 'a'] > 1) {
+            ans[l] = ans[r] = curr;
+            freq[curr - 'a'] -= 2;
+            if (func(i + 1, target, ans, l + 1, r - 1, freq, end)) {
+                return true;
+            }
+            freq[curr - 'a'] += 2;
+        }
+        if (next != '1') {
+            ans[l] = ans[r] = next;
+            freq[next - 'a'] -= 2;
+            if (func(i + 1, target, ans, l + 1, r - 1, freq, true)) {
+                return true;
+            }
+        }
+        ans[l] = ans[r] = '#';
+        return false;
+    }
+
+    public String lexPalindromicPermutation(String s, String target) {
+        int[] freq = new int[26];
+        for (int i = 0; i < s.length(); i++) {
+            freq[s.charAt(i) - 'a']++;
+        }
+        int oddc = 0;
+        for (int i = 0; i < 26; i++) {
+            if (freq[i] % 2 == 1) {
+                oddc++;
+            }
+        }
+        if (oddc > 1) {
+            return "";
+        }
+        char[] ans = new char[s.length()];
+        Arrays.fill(ans, '#');
+        if (func(0, target, ans, 0, s.length() - 1, freq, false)) {
+            return new String(ans);
+        }
+        return "";
+    }
+}