Added tasks 3692-3700

Author: javadev

src/main/kotlin/g3601_3700/s3692_majority_frequency_characters/Solution.kt

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+package g3601_3700.s3692_majority_frequency_characters
+
+// #Easy #Biweekly_Contest_166 #2025_10_03_Time_2_ms_(100.00%)_Space_43.05_MB_(100.00%)
+
+class Solution {
+    fun majorityFrequencyGroup(s: String): String {
+        val cntArray = IntArray(26)
+        for (i in 0..<s.length) {
+            cntArray[s[i].code - 'a'.code]++
+        }
+        val freq = IntArray(s.length + 1)
+        for (i in 0..25) {
+            if (cntArray[i] > 0) {
+                freq[cntArray[i]]++
+            }
+        }
+        var size = 0
+        var bfreq = 0
+        for (i in 0..s.length) {
+            val si = freq[i]
+            if (si > size || (si == size && i > bfreq)) {
+                size = si
+                bfreq = i
+            }
+        }
+        val sb = StringBuilder()
+        for (i in 0..25) {
+            if (cntArray[i] == bfreq) {
+                sb.append((i + 'a'.code).toChar())
+            }
+        }
+        return sb.toString()
+    }
+}

src/main/kotlin/g3601_3700/s3692_majority_frequency_characters/readme.md

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+3692\. Majority Frequency Characters
+
+Easy
+
+You are given a string `s` consisting of lowercase English letters.
+
+The **frequency group** for a value `k` is the set of characters that appear exactly `k` times in s.
+
+The **majority frequency group** is the frequency group that contains the largest number of **distinct** characters.
+
+Return a string containing all characters in the majority frequency group, in **any** order. If two or more frequency groups tie for that largest size, pick the group whose frequency `k` is **larger**.
+
+**Example 1:**
+
+**Input:** s = "aaabbbccdddde"
+
+**Output:** "ab"
+
+**Explanation:**
+
+| Frequency (k) | Distinct characters in group | Group size | Majority? |
+|---------------|------------------------------|------------|-----------|
+| 4             | {d}                          | 1          | No        |
+| 3             | {a, b}                       | 2          | **Yes**   |
+| 2             | {c}                          | 1          | No        |
+| 1             | {e}                          | 1          | No        |
+
+Both characters `'a'` and `'b'` share the same frequency 3, they are in the majority frequency group. `"ba"` is also a valid answer.
+
+**Example 2:**
+
+**Input:** s = "abcd"
+
+**Output:** "abcd"
+
+**Explanation:**
+
+| Frequency (k) | Distinct characters in group | Group size | Majority? |
+|---------------|------------------------------|------------|-----------|
+| 1             | {a, b, c, d}                 | 4          | **Yes**   |
+
+All characters share the same frequency 1, they are all in the majority frequency group.
+
+**Example 3:**
+
+**Input:** s = "pfpfgi"
+
+**Output:** "fp"
+
+**Explanation:**
+
+| Frequency (k) | Distinct characters in group | Group size | Majority?                        |
+|---------------|------------------------------|------------|----------------------------------|
+| 2             | {p, f}                       | 2          | **Yes**                          |
+| 1             | {g, i}                       | 2          | No (tied size, lower frequency)  |
+
+Both characters `'p'` and `'f'` share the same frequency 2, they are in the majority frequency group. There is a tie in group size with frequency 1, but we pick the higher frequency: 2.
+
+**Constraints:**
+
+*   `1 <= s.length <= 100`
+*   `s` consists only of lowercase English letters.

src/main/kotlin/g3601_3700/s3693_climbing_stairs_ii/Solution.kt

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+package g3601_3700.s3693_climbing_stairs_ii
+
+// #Medium #Biweekly_Contest_166 #2025_10_03_Time_8_ms_(100.00%)_Space_80.61_MB_(12.90%)
+
+import kotlin.math.min
+
+@Suppress("unused")
+class Solution {
+    fun climbStairs(n: Int, costs: IntArray): Int {
+        if (costs.size == 1) {
+            return costs[0] + 1
+        }
+        var one = costs[0] + 1
+        var two = min(one + costs[1] + 1, costs[1] + 4)
+        if (costs.size < 3) {
+            return two
+        }
+        var three = min(one + costs[2] + 4, min(two + costs[2] + 1, costs[2] + 9))
+        if (costs.size < 4) {
+            return three
+        }
+        for (i in 3..<costs.size) {
+            val four =
+                (
+                    min(
+                        three + costs[i] + 1,
+                        min(two + costs[i] + 4, one + costs[i] + 9),
+                    )
+                    )
+            one = two
+            two = three
+            three = four
+        }
+        return three
+    }
+}

src/main/kotlin/g3601_3700/s3693_climbing_stairs_ii/readme.md

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+3693\. Climbing Stairs II
+
+Medium
+
+You are climbing a staircase with `n + 1` steps, numbered from 0 to `n`.
+
+You are also given a **1-indexed** integer array `costs` of length `n`, where `costs[i]` is the cost of step `i`.
+
+From step `i`, you can jump **only** to step `i + 1`, `i + 2`, or `i + 3`. The cost of jumping from step `i` to step `j` is defined as: <code>costs[j] + (j - i)<sup>2</sup></code>
+
+You start from step 0 with `cost = 0`.
+
+Return the **minimum** total cost to reach step `n`.
+
+**Example 1:**
+
+**Input:** n = 4, costs = [1,2,3,4]
+
+**Output:** 13
+
+**Explanation:**
+
+One optimal path is `0 → 1 → 2 → 4`
+
+Jump
+
+Cost Calculation
+
+Cost
+
+0 → 1
+
+<code>costs[1] + (1 - 0)<sup>2</sup> = 1 + 1</code>
+
+2
+
+1 → 2
+
+<code>costs[2] + (2 - 1)<sup>2</sup> = 2 + 1</code>
+
+3
+
+2 → 4
+
+<code>costs[4] + (4 - 2)<sup>2</sup> = 4 + 4</code>
+
+8
+
+Thus, the minimum total cost is `2 + 3 + 8 = 13`
+
+**Example 2:**
+
+**Input:** n = 4, costs = [5,1,6,2]
+
+**Output:** 11
+
+**Explanation:**
+
+One optimal path is `0 → 2 → 4`
+
+Jump
+
+Cost Calculation
+
+Cost
+
+0 → 2
+
+<code>costs[2] + (2 - 0)<sup>2</sup> = 1 + 4</code>
+
+5
+
+2 → 4
+
+<code>costs[4] + (4 - 2)<sup>2</sup> = 2 + 4</code>
+
+6
+
+Thus, the minimum total cost is `5 + 6 = 11`
+
+**Example 3:**
+
+**Input:** n = 3, costs = [9,8,3]
+
+**Output:** 12
+
+**Explanation:**
+
+The optimal path is `0 → 3` with total cost = <code>costs[3] + (3 - 0)<sup>2</sup> = 3 + 9 = 12</code>
+
+**Constraints:**
+
+*   <code>1 <= n == costs.length <= 10<sup>5</sup></code>
+*   <code>1 <= costs[i] <= 10<sup>4</sup></code>

src/main/kotlin/g3601_3700/s3694_distinct_points_reachable_after_substring_removal/Solution.kt

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+package g3601_3700.s3694_distinct_points_reachable_after_substring_removal
+
+// #Medium #Biweekly_Contest_166 #2025_10_03_Time_46_ms_(100.00%)_Space_48.62_MB_(100.00%)
+
+class Solution {
+    fun distinctPoints(s: String, k: Int): Int {
+        val seen: MutableSet<Long> = HashSet()
+        seen.add(0L)
+        var x = 0
+        var y = 0
+        for (i in k..<s.length) {
+            // add new step
+            when (s[i]) {
+                'U' -> y++
+                'D' -> y--
+                'L' -> x++
+                'R' -> x--
+                else -> x--
+            }
+            // remove old step
+            when (s[i - k]) {
+                'U' -> y--
+                'D' -> y++
+                'L' -> x--
+                'R' -> x++
+                else -> x++
+            }
+            seen.add(1000000L * x + y)
+        }
+        return seen.size
+    }
+}

src/main/kotlin/g3601_3700/s3694_distinct_points_reachable_after_substring_removal/readme.md

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+3694\. Distinct Points Reachable After Substring Removal
+
+Medium
+
+You are given a string `s` consisting of characters `'U'`, `'D'`, `'L'`, and `'R'`, representing moves on an infinite 2D Cartesian grid.
+
+*   `'U'`: Move from `(x, y)` to `(x, y + 1)`.
+*   `'D'`: Move from `(x, y)` to `(x, y - 1)`.
+*   `'L'`: Move from `(x, y)` to `(x - 1, y)`.
+*   `'R'`: Move from `(x, y)` to `(x + 1, y)`.
+
+You are also given a positive integer `k`.
+
+You **must** choose and remove **exactly one** contiguous substring of length `k` from `s`. Then, start from coordinate `(0, 0)` and perform the remaining moves in order.
+
+Return an integer denoting the number of **distinct** final coordinates reachable.
+
+**Example 1:**
+
+**Input:** s = "LUL", k = 1
+
+**Output:** 2
+
+**Explanation:**
+
+After removing a substring of length 1, `s` can be `"UL"`, `"LL"` or `"LU"`. Following these moves, the final coordinates will be `(-1, 1)`, `(-2, 0)` and `(-1, 1)` respectively. There are two distinct points `(-1, 1)` and `(-2, 0)` so the answer is 2.
+
+**Example 2:**
+
+**Input:** s = "UDLR", k = 4
+
+**Output:** 1
+
+**Explanation:**
+
+After removing a substring of length 4, `s` can only be the empty string. The final coordinates will be `(0, 0)`. There is only one distinct point `(0, 0)` so the answer is 1.
+
+**Example 3:**
+
+**Input:** s = "UU", k = 1
+
+**Output:** 1
+
+**Explanation:**
+
+After removing a substring of length 1, `s` becomes `"U"`, which always ends at `(0, 1)`, so there is only one distinct final coordinate.
+
+**Constraints:**
+
+*   <code>1 <= s.length <= 10<sup>5</sup></code>
+*   `s` consists of only `'U'`, `'D'`, `'L'`, and `'R'`.
+*   `1 <= k <= s.length`

src/main/kotlin/g3601_3700/s3695_maximize_alternating_sum_using_swaps/Solution.kt

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+package g3601_3700.s3695_maximize_alternating_sum_using_swaps
+
+// #Hard #Biweekly_Contest_166 #2025_10_03_Time_61_ms_(100.00%)_Space_105.29_MB_(100.00%)
+
+class Solution {
+    private lateinit var root: IntArray
+
+    fun maxAlternatingSum(nums: IntArray, swaps: Array<IntArray>): Long {
+        val n = nums.size
+        root = IntArray(n) { it }
+        val list = Array(n) { ArrayList<Int>() }
+        val oddCount = IntArray(n)
+        for (s in swaps) {
+            union(s[0], s[1])
+        }
+        for (i in nums.indices) {
+            val r = findRoot(i)
+            list[r].add(nums[i])
+            if (i % 2 == 1) {
+                oddCount[r]++
+            }
+        }
+
+        var result = 0L
+        for (i in 0 until n) {
+            if (root[i] != i) {
+                continue
+            }
+            val currentList = list[i]
+            val currentOddCount = oddCount[i]
+            currentList.sort()
+            for (j in currentList.indices) {
+                val value = currentList[j].toLong()
+                val multiplier = if (j < currentOddCount) -1 else 1
+                result += value * multiplier
+            }
+        }
+        return result
+    }
+
+    private fun union(a: Int, b: Int) {
+        val rootA = findRoot(a)
+        val rootB = findRoot(b)
+        if (rootA != rootB) {
+            if (rootA < rootB) {
+                root[rootB] = rootA
+            } else {
+                root[rootA] = rootB
+            }
+        }
+    }
+
+    private fun findRoot(a: Int): Int {
+        if (a == root[a]) {
+            return a
+        }
+        return findRoot(root[a]).also { root[a] = it }
+    }
+}