Added tasks 3232-3235

Author: javadev

README.md

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+package g3201_3300.s3232_find_if_digit_game_can_be_won
+
+// #Easy #Array #Math #2024_08_03_Time_194_ms_(36.00%)_Space_40_MB_(5.33%)
+
+class Solution {
+    fun canAliceWin(nums: IntArray): Boolean {
+        var sdSum = 0
+        var ddSum = 0
+        for (num in nums) {
+            if (num / 10 == 0) {
+                sdSum += num
+            } else {
+                ddSum += num
+            }
+        }
+        return sdSum != ddSum
+    }
+}

src/main/kotlin/g3201_3300/s3232_find_if_digit_game_can_be_won/Solution.kt

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+3232\. Find if Digit Game Can Be Won
+
+Easy
+
+You are given an array of **positive** integers `nums`.
+
+Alice and Bob are playing a game. In the game, Alice can choose **either** all single-digit numbers or all double-digit numbers from `nums`, and the rest of the numbers are given to Bob. Alice wins if the sum of her numbers is **strictly greater** than the sum of Bob's numbers.
+
+Return `true` if Alice can win this game, otherwise, return `false`.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,4,10]
+
+**Output:** false
+
+**Explanation:**
+
+Alice cannot win by choosing either single-digit or double-digit numbers.
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,4,5,14]
+
+**Output:** true
+
+**Explanation:**
+
+Alice can win by choosing single-digit numbers which have a sum equal to 15.
+
+**Example 3:**
+
+**Input:** nums = [5,5,5,25]
+
+**Output:** true
+
+**Explanation:**
+
+Alice can win by choosing double-digit numbers which have a sum equal to 25.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `1 <= nums[i] <= 99`

src/main/kotlin/g3201_3300/s3232_find_if_digit_game_can_be_won/readme.md

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+package g3201_3300.s3233_find_the_count_of_numbers_which_are_not_special
+
+// #Medium #Array #Math #Number_Theory #2024_08_03_Time_215_ms_(76.19%)_Space_36.9_MB_(61.90%)
+
+import kotlin.math.sqrt
+
+class Solution {
+    fun nonSpecialCount(l: Int, r: Int): Int {
+        val primes = sieveOfEratosthenes(sqrt(r.toDouble()).toInt())
+        var specialCount = 0
+
+        for (prime in primes) {
+            val primeSquare = prime.toLong() * prime
+            if (primeSquare in l..r) {
+                specialCount++
+            }
+        }
+
+        val totalNumbersInRange = r - l + 1
+        return totalNumbersInRange - specialCount
+    }
+
+    private fun sieveOfEratosthenes(n: Int): List<Int> {
+        val isPrime = BooleanArray(n + 1)
+        for (i in 2..n) {
+            isPrime[i] = true
+        }
+
+        var p = 2
+        while (p * p <= n) {
+            if (isPrime[p]) {
+                var i = p * p
+                while (i <= n) {
+                    isPrime[i] = false
+                    i += p
+                }
+            }
+            p++
+        }
+
+        val primes: MutableList<Int> = ArrayList()
+        for (i in 2..n) {
+            if (isPrime[i]) {
+                primes.add(i)
+            }
+        }
+        return primes
+    }
+}

src/main/kotlin/g3201_3300/s3233_find_the_count_of_numbers_which_are_not_special/Solution.kt

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+3233\. Find the Count of Numbers Which Are Not Special
+
+Medium
+
+You are given 2 **positive** integers `l` and `r`. For any number `x`, all positive divisors of `x` _except_ `x` are called the **proper divisors** of `x`.
+
+A number is called **special** if it has exactly 2 **proper divisors**. For example:
+
+*   The number 4 is _special_ because it has proper divisors 1 and 2.
+*   The number 6 is _not special_ because it has proper divisors 1, 2, and 3.
+
+Return the count of numbers in the range `[l, r]` that are **not** **special**.
+
+**Example 1:**
+
+**Input:** l = 5, r = 7
+
+**Output:** 3
+
+**Explanation:**
+
+There are no special numbers in the range `[5, 7]`.
+
+**Example 2:**
+
+**Input:** l = 4, r = 16
+
+**Output:** 11
+
+**Explanation:**
+
+The special numbers in the range `[4, 16]` are 4 and 9.
+
+**Constraints:**
+
+*   <code>1 <= l <= r <= 10<sup>9</sup></code>

src/main/kotlin/g3201_3300/s3233_find_the_count_of_numbers_which_are_not_special/readme.md

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+package g3201_3300.s3234_count_the_number_of_substrings_with_dominant_ones
+
+// #Medium #String #Sliding_Window #Enumeration
+// #2024_08_03_Time_356_ms_(100.00%)_Space_38_MB_(76.92%)
+
+import kotlin.math.min
+
+class Solution {
+    fun numberOfSubstrings(s: String): Int {
+        val zero: MutableList<Int> = ArrayList()
+        zero.add(-1)
+        var result = 0
+        for (i in s.indices) {
+            if (s[i] == '0') {
+                zero.add(i)
+            } else {
+                result += i - zero[zero.size - 1]
+            }
+            for (j in 1 until zero.size) {
+                val len = j * (j + 1)
+                if (len > i + 1) {
+                    break
+                }
+                val prev = zero[zero.size - j - 1]
+                val from = min((i - len + 1), zero[zero.size - j])
+                if (from > prev) {
+                    result += from - prev
+                }
+            }
+        }
+        return result
+    }
+}

src/main/kotlin/g3201_3300/s3234_count_the_number_of_substrings_with_dominant_ones/Solution.kt

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+3234\. Count the Number of Substrings With Dominant Ones
+
+Medium
+
+You are given a binary string `s`.
+
+Return the number of substrings with **dominant** ones.
+
+A string has **dominant** ones if the number of ones in the string is **greater than or equal to** the **square** of the number of zeros in the string.
+
+**Example 1:**
+
+**Input:** s = "00011"
+
+**Output:** 5
+
+**Explanation:**
+
+The substrings with dominant ones are shown in the table below.
+
+| i | j | s[i..j] | Number of Zeros | Number of Ones |
+|---|---|---------|-----------------|----------------|
+| 3 | 3 | 1       | 0               | 1              |
+| 4 | 4 | 1       | 0               | 1              |
+| 2 | 3 | 01      | 1               | 1              |
+| 3 | 4 | 11      | 0               | 2              |
+| 2 | 4 | 011     | 1               | 2              |
+
+**Example 2:**
+
+**Input:** s = "101101"
+
+**Output:** 16
+
+**Explanation:**
+
+The substrings with **non-dominant** ones are shown in the table below.
+
+Since there are 21 substrings total and 5 of them have non-dominant ones, it follows that there are 16 substrings with dominant ones.
+
+| i | j | s[i..j] | Number of Zeros | Number of Ones |
+|---|---|---------|-----------------|----------------|
+| 1 | 1 | 0       | 1               | 0              |
+| 4 | 4 | 0       | 1               | 0              |
+| 1 | 4 | 0110    | 2               | 2              |
+| 0 | 4 | 10110   | 2               | 3              |
+| 1 | 5 | 01101   | 2               | 3              |
+
+**Constraints:**
+
+*   <code>1 <= s.length <= 4 * 10<sup>4</sup></code>
+*   `s` consists only of characters `'0'` and `'1'`.

src/main/kotlin/g3201_3300/s3234_count_the_number_of_substrings_with_dominant_ones/readme.md

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+package g3201_3300.s3235_check_if_the_rectangle_corner_is_reachable
+
+// #Hard #Array #Math #Depth_First_Search #Breadth_First_Search #Union_Find #Geometry
+// #2024_08_03_Time_612_ms_(66.67%)_Space_50.5_MB_(88.89%)
+
+import kotlin.math.pow
+import kotlin.math.sqrt
+
+class Solution {
+    fun canReachCorner(x: Int, y: Int, circles: Array<IntArray>): Boolean {
+        val n = circles.size
+        val ds = DisjointSet(n + 5)
+
+        // Special nodes for boundaries
+        val leftBoundary = n + 3
+        val topBoundary = n
+        val rightBoundary = n + 1
+        val bottomBoundary = n + 2
+
+        var i = 0
+        for (it in circles) {
+            val xi = it[0]
+            val yi = it[1]
+            val ri = it[2]
+            if (yi - ri >= y || xi - ri >= x) {
+                continue
+            }
+            if (((xi > (x + y) || yi > y) && (xi > x || yi > x + y))) {
+                continue
+            }
+            if (xi <= ri) {
+                ds.dsu(i, leftBoundary)
+            }
+            if (yi <= ri) {
+                ds.dsu(i, topBoundary)
+            }
+            if (x - xi <= ri) {
+                ds.dsu(i, rightBoundary)
+            }
+            if (y - yi <= ri) {
+                ds.dsu(i, bottomBoundary)
+            }
+            i++
+        }
+
+        // Union circles that overlap
+        i = 0
+        while (i < n) {
+            val x1 = circles[i][0]
+            val y1 = circles[i][1]
+            val r1 = circles[i][2]
+            if (y1 - r1 >= y || x1 - r1 >= x) {
+                i++
+                continue
+            }
+            if (((x1 > (x + y) || y1 > y) && (x1 > x || y1 > x + y))) {
+                i++
+                continue
+            }
+
+            for (j in i + 1 until n) {
+                val x2 = circles[j][0]
+                val y2 = circles[j][1]
+                val r2 = circles[j][2]
+                val dist = sqrt(
+                    (x1 - x2.toDouble()).pow(2.0) + (y1 - y2.toDouble()).pow(2.0)
+                )
+                if (dist <= (r1 + r2)) {
+                    ds.dsu(i, j)
+                }
+            }
+            i++
+        }
+
+        // Check if left is connected to right or top is connected to bottom
+        if (ds.findUpar(leftBoundary) == ds.findUpar(rightBoundary) ||
+            ds.findUpar(leftBoundary) == ds.findUpar(topBoundary)
+        ) {
+            return false
+        }
+        return (
+            ds.findUpar(bottomBoundary) != ds.findUpar(rightBoundary) &&
+                ds.findUpar(bottomBoundary) != ds.findUpar(topBoundary)
+            )
+    }
+
+    private class DisjointSet(n: Int) {
+        private val parent: IntArray
+        private val size = IntArray(n + 1)
+
+        init {
+            size.fill(1)
+            parent = IntArray(n + 1)
+            for (i in 0..n) {
+                parent[i] = i
+            }
+        }
+
+        fun findUpar(u: Int): Int {
+            if (u == parent[u]) {
+                return u
+            }
+            parent[u] = findUpar(parent[u])
+            return parent[u]
+        }
+
+        fun dsu(u: Int, v: Int) {
+            val ulpu = findUpar(u)
+            val ulpv = findUpar(v)
+            if (ulpv == ulpu) {
+                return
+            }
+            if (size[ulpu] < size[ulpv]) {
+                parent[ulpu] = ulpv
+                size[ulpv] += size[ulpu]
+            } else {
+                parent[ulpv] = ulpu
+                size[ulpu] += size[ulpv]
+            }
+        }
+    }
+}