Added tasks 3737-3747

src/main/java/g3701_3800/s3737_count_subarrays_with_majority_element_i/Solution.java

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+package g3701_3800.s3737_count_subarrays_with_majority_element_i;
+
+// #Medium #Array #Hash_Table #Prefix_Sum #Counting #Divide_and_Conquer #Segment_Tree #Merge_Sort
+// #Senior #Biweekly_Contest_169 #2026_04_26_Time_1_ms_(100.00%)_Space_46.96_MB_(57.62%)
+
+public class Solution {
+    public int countMajoritySubarrays(int[] a, int target) {
+        int n = a.length;
+        int pre = n + 1;
+        int res = 0;
+        int[] count = new int[2 * n + 2];
+        int[] acc = new int[2 * n + 2];
+        count[pre] = acc[pre] = 1;
+        for (int i : a) {
+            pre += (i == target ? 1 : -1);
+            count[pre]++;
+            acc[pre] = acc[pre - 1] + count[pre];
+            res += acc[pre - 1];
+        }
+        return res;
+    }
+}

src/main/java/g3701_3800/s3737_count_subarrays_with_majority_element_i/readme.md

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+3737\. Count Subarrays With Majority Element I
+
+Medium
+
+You are given an integer array `nums` and an integer `target`.
+
+Return the number of **non-empty subarrays** of `nums` in which `target` is the **majority element**.
+
+The **majority element** of a subarray is the element that appears **strictly** **more than half** of the times in that subarray.
+
+**Example 1:**
+
+**Input:** nums = [1,2,2,3], target = 2
+
+**Output:** 5
+
+**Explanation:**
+
+Valid subarrays with `target = 2` as the majority element:
+
+*   `nums[1..1] = [2]`
+*   `nums[2..2] = [2]`
+*   `nums[1..2] = [2,2]`
+*   `nums[0..2] = [1,2,2]`
+*   `nums[1..3] = [2,2,3]`
+
+So there are 5 such subarrays.
+
+**Example 2:**
+
+**Input:** nums = [1,1,1,1], target = 1
+
+**Output:** 10
+
+**Explanation:**
+
+All 10 subarrays have 1 as the majority element.
+
+**Example 3:**
+
+**Input:** nums = [1,2,3], target = 4
+
+**Output:** 0
+
+**Explanation:**
+
+`target = 4` does not appear in `nums` at all. Therefore, there cannot be any subarray where 4 is the majority element. Hence the answer is 0.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 1000`
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>
+*   <code>1 <= target <= 10<sup>9</sup></code>

src/main/java/g3701_3800/s3738_longest_non_decreasing_subarray_after_replacing_at_most_one_element/Solution.java

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+package g3701_3800.s3738_longest_non_decreasing_subarray_after_replacing_at_most_one_element;
+
+// #Medium #Array #Dynamic_Programming #Staff #Biweekly_Contest_169
+// #2026_04_26_Time_7_ms_(99.28%)_Space_123.88_MB_(5.07%)
+
+public class Solution {
+    public int longestSubarray(int[] nums) {
+        int n = nums.length;
+        if (n < 3) {
+            return n;
+        }
+        int i = 0;
+        int len1 = 0;
+        int len2 = 0;
+        int ans = 2;
+        while (i < n) {
+            int l = (i == 0) ? -1000 * 1000 * 10 : nums[i - 1];
+            int r = (i == n - 1) ? 1000 * 1000 * 10 : nums[i + 1];
+            if (l <= nums[i]) {
+                len2++;
+                ans = Math.max(len2 + 1, ans);
+            } else if (r >= nums[i]) {
+                int j = i;
+                len1 = len2;
+                while (j < n - 1 && nums[j] <= nums[j + 1]) {
+                    j++;
+                }
+                len2 = j - i + 1;
+                ans = Math.max(len2 + 1, ans);
+                if (l <= r || (len1 > 1 && nums[i - 2] <= nums[i])) {
+                    ans = Math.max(len1 + len2, ans);
+                }
+                i = j;
+            } else {
+                len2 = 0;
+            }
+            i++;
+        }
+        ans = Math.min(ans, n);
+        return ans;
+    }
+}

src/main/java/g3701_3800/s3738_longest_non_decreasing_subarray_after_replacing_at_most_one_element/readme.md

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+3738\. Longest Non-Decreasing Subarray After Replacing at Most One Element
+
+Medium
+
+You are given an integer array `nums`.
+
+You are allowed to replace **at most** one element in the array with any other integer value of your choice.
+
+Return the length of the **longest non-decreasing subarray** that can be obtained after performing at most one replacement.
+
+An array is said to be **non-decreasing** if each element is greater than or equal to its previous one (if it exists).
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,1,2]
+
+**Output:** 4
+
+**Explanation:**
+
+Replacing `nums[3] = 1` with 3 gives the array [1, 2, 3, 3, 2].
+
+The longest non-decreasing subarray is [1, 2, 3, 3], which has a length of 4.
+
+**Example 2:**
+
+**Input:** nums = [2,2,2,2,2]
+
+**Output:** 5
+
+**Explanation:**
+
+All elements in `nums` are equal, so it is already non-decreasing and the entire `nums` forms a subarray of length 5.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>

src/main/java/g3701_3800/s3739_count_subarrays_with_majority_element_ii/Solution.java

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+package g3701_3800.s3739_count_subarrays_with_majority_element_ii;
+
+// #Hard #Array #Hash_Table #Prefix_Sum #Divide_and_Conquer #Segment_Tree #Merge_Sort #Senior_Staff
+// #Biweekly_Contest_169 #2026_04_26_Time_3_ms_(100.00%)_Space_90.94_MB_(36.19%)
+
+public class Solution {
+    public long countMajoritySubarrays(int[] nums, int target) {
+        int n = nums.length;
+        int pre = n + 1;
+        long[] count = new long[2 * n + 2];
+        long[] acc = new long[2 * n + 2];
+        long res = 0;
+        count[pre] = acc[pre] = 1;
+        for (int a : nums) {
+            pre += (a == target ? 1 : -1);
+            acc[pre] = ++count[pre] + acc[pre - 1];
+            res += acc[pre - 1];
+        }
+        return res;
+    }
+}

src/main/java/g3701_3800/s3739_count_subarrays_with_majority_element_ii/readme.md

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+3739\. Count Subarrays With Majority Element II
+
+Hard
+
+You are given an integer array `nums` and an integer `target`.
+
+Return the number of **non-empty subarrays** of `nums` in which `target` is the **majority element**.
+
+The **majority element** of a subarray is the element that appears **strictly more than half** of the times in that subarray.
+
+**Example 1:**
+
+**Input:** nums = [1,2,2,3], target = 2
+
+**Output:** 5
+
+**Explanation:**
+
+Valid subarrays with `target = 2` as the majority element:
+
+*   `nums[1..1] = [2]`
+*   `nums[2..2] = [2]`
+*   `nums[1..2] = [2,2]`
+*   `nums[0..2] = [1,2,2]`
+*   `nums[1..3] = [2,2,3]`
+
+So there are 5 such subarrays.
+
+**Example 2:**
+
+**Input:** nums = [1,1,1,1], target = 1
+
+**Output:** 10
+
+**Explanation:**
+
+All 10 subarrays have 1 as the majority element.
+
+**Example 3:**
+
+**Input:** nums = [1,2,3], target = 4
+
+**Output:** 0
+
+**Explanation:**
+
+`target = 4` does not appear in `nums` at all. Therefore, there cannot be any subarray where 4 is the majority element. Hence the answer is 0.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>
+*   <code>1 <= target <= 10<sup>9</sup></code>

src/main/java/g3701_3800/s3740_minimum_distance_between_three_equal_elements_i/Solution.java

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+package g3701_3800.s3740_minimum_distance_between_three_equal_elements_i;
+
+// #Easy #Array #Hash_Table #Mid_Level #Weekly_Contest_475
+// #2026_04_26_Time_1_ms_(99.99%)_Space_44.21_MB_(75.36%)
+
+public class Solution {
+    public int minimumDistance(int[] nums) {
+        int len = nums.length;
+        int[] last2 = new int[len];
+        int res = 200;
+        for (int i = 0; i < len; i++) {
+            int val = nums[i] - 1;
+            int pos = i + 1;
+            int pack = last2[val];
+            int old = pack & 255;
+            int cur = pack >> 8;
+            last2[val] = cur | (pos << 8);
+            if (old > 0) {
+                res = Math.min(res, (pos - old) << 1);
+            }
+        }
+        return res == 200 ? -1 : res;
+    }
+}

src/main/java/g3701_3800/s3740_minimum_distance_between_three_equal_elements_i/readme.md

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+3740\. Minimum Distance Between Three Equal Elements I
+
+Easy
+
+You are given an integer array `nums`.
+
+A tuple `(i, j, k)` of 3 **distinct** indices is **good** if `nums[i] == nums[j] == nums[k]`.
+
+The **distance** of a **good** tuple is `abs(i - j) + abs(j - k) + abs(k - i)`, where `abs(x)` denotes the **absolute value** of `x`.
+
+Return an integer denoting the **minimum** possible **distance** of a **good** tuple. If no **good** tuples exist, return `-1`.
+
+**Example 1:**
+
+**Input:** nums = [1,2,1,1,3]
+
+**Output:** 6
+
+**Explanation:**
+
+The minimum distance is achieved by the good tuple `(0, 2, 3)`.
+
+`(0, 2, 3)` is a good tuple because `nums[0] == nums[2] == nums[3] == 1`. Its distance is `abs(0 - 2) + abs(2 - 3) + abs(3 - 0) = 2 + 1 + 3 = 6`.
+
+**Example 2:**
+
+**Input:** nums = [1,1,2,3,2,1,2]
+
+**Output:** 8
+
+**Explanation:**
+
+The minimum distance is achieved by the good tuple `(2, 4, 6)`.
+
+`(2, 4, 6)` is a good tuple because `nums[2] == nums[4] == nums[6] == 2`. Its distance is `abs(2 - 4) + abs(4 - 6) + abs(6 - 2) = 2 + 2 + 4 = 8`.
+
+**Example 3:**
+
+**Input:** nums = [1]
+
+**Output:** \-1
+
+**Explanation:**
+
+There are no good tuples. Therefore, the answer is -1.
+
+**Constraints:**
+
+*   `1 <= n == nums.length <= 100`
+*   `1 <= nums[i] <= n`

src/main/java/g3701_3800/s3741_minimum_distance_between_three_equal_elements_ii/Solution.java

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+package g3701_3800.s3741_minimum_distance_between_three_equal_elements_ii;
+
+// #Medium #Array #Hash_Table #Senior #Weekly_Contest_475
+// #2026_04_26_Time_6_ms_(99.60%)_Space_161.24_MB_(96.23%)
+
+public class Solution {
+    public int minimumDistance(int[] nums) {
+        int n = nums.length;
+        int ans = Integer.MAX_VALUE;
+        int[] prev1 = new int[n + 1];
+        int[] prev2 = new int[n + 1];
+        for (int i = 0; i < n + 1; i++) {
+            prev1[i] = prev2[i] = -1;
+        }
+        for (int i = 0; i < n; i++) {
+            int value = nums[i];
+            if (prev2[value] != -1) {
+                ans = Math.min(ans, (i - prev2[value]));
+            }
+            prev2[value] = prev1[value];
+            prev1[value] = i;
+        }
+        if (ans < 100002) {
+            return ans * 2;
+        }
+        return -1;
+    }
+}

src/main/java/g3701_3800/s3741_minimum_distance_between_three_equal_elements_ii/readme.md

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+3741\. Minimum Distance Between Three Equal Elements II
+
+Medium
+
+You are given an integer array `nums`.
+
+A tuple `(i, j, k)` of 3 **distinct** indices is **good** if `nums[i] == nums[j] == nums[k]`.
+
+The **distance** of a **good** tuple is `abs(i - j) + abs(j - k) + abs(k - i)`, where `abs(x)` denotes the **absolute value** of `x`.
+
+Return an integer denoting the **minimum** possible **distance** of a **good** tuple. If no **good** tuples exist, return `-1`.
+
+**Example 1:**
+
+**Input:** nums = [1,2,1,1,3]
+
+**Output:** 6
+
+**Explanation:**
+
+The minimum distance is achieved by the good tuple `(0, 2, 3)`.
+
+`(0, 2, 3)` is a good tuple because `nums[0] == nums[2] == nums[3] == 1`. Its distance is `abs(0 - 2) + abs(2 - 3) + abs(3 - 0) = 2 + 1 + 3 = 6`.
+
+**Example 2:**
+
+**Input:** nums = [1,1,2,3,2,1,2]
+
+**Output:** 8
+
+**Explanation:**
+
+The minimum distance is achieved by the good tuple `(2, 4, 6)`.
+
+`(2, 4, 6)` is a good tuple because `nums[2] == nums[4] == nums[6] == 2`. Its distance is `abs(2 - 4) + abs(4 - 6) + abs(6 - 2) = 2 + 2 + 4 = 8`.
+
+**Example 3:**
+
+**Input:** nums = [1]
+
+**Output:** \-1
+
+**Explanation:**
+
+There are no good tuples. Therefore, the answer is -1.
+
+**Constraints:**
+
+*   <code>1 <= n == nums.length <= 10<sup>5</sup></code>
+*   `1 <= nums[i] <= n`