diff --git a/contains-duplicate/jdalma.kt b/contains-duplicate/jdalma.kt
new file mode 100644
index 000000000..d4684df3f
--- /dev/null
+++ b/contains-duplicate/jdalma.kt
@@ -0,0 +1,65 @@
+package leetcode_study
+
+import io.kotest.matchers.equals.shouldBeEqual
+import org.junit.jupiter.api.Test
+
+class `contains-duplicate`{
+
+    fun containsDuplicate(nums: IntArray): Boolean {
+        return usingSet(nums)
+    }
+
+    // 1. 중첩 반복문 - 시간초과
+    // 시간복잡도: O(n^2), 공간복잡도: O(1)
+    private fun usingNestedLoop(nums: IntArray): Boolean {
+        nums.forEachIndexed { i, e1 ->
+            nums.forEachIndexed { j, e2 ->
+                if (i != j && e1 == e2) {
+                    return true
+                }
+            }
+        }
+        return false
+    }
+
+    // 2. 정렬 후 순회
+    // 시간복잡도:  O(n * log(n)), 공간복잡도: O(1)
+    private fun usingSort(nums: IntArray): Boolean {
+        nums.sort() // DualPivotQuicksort -> O(n log(n))
+        for (index in 1 until nums.size) {
+            val prev = nums[index - 1]
+            val curr = nums[index]
+            if (prev == curr) {
+                return true
+            }
+        }
+        return false
+    }
+
+    // 3. 자료구조 Set 사용
+    // 시간복잡도: O(n), 공간복잡도: O(n)
+    private fun usingSet(nums: IntArray): Boolean {
+        val set = nums.toSet()
+        return nums.size != set.size
+    }
+
+    @Test
+    fun 동일한_원소가_존재하면_true를_반환한다() {
+        val nums1 = intArrayOf(1, 2, 3, 1)
+        val nums2 = intArrayOf(1, 2, 3, 2)
+        val nums3 = intArrayOf(1, 1, 1, 1, 3, 3, 4, 3, 2, 4, 2)
+
+        containsDuplicate(nums1) shouldBeEqual true
+        containsDuplicate(nums2) shouldBeEqual true
+        containsDuplicate(nums3) shouldBeEqual true
+    }
+
+    @Test
+    fun 동일한_원소가_존재하지_않으면_false를_반환한다() {
+        val nums1 = intArrayOf(1, 2, 3, 4)
+        val nums2 = intArrayOf(1, 5, 7)
+
+        containsDuplicate(nums1) shouldBeEqual false
+        containsDuplicate(nums2) shouldBeEqual false
+    }
+}
diff --git a/kth-smallest-element-in-a-bst/jdalma.kt b/kth-smallest-element-in-a-bst/jdalma.kt
new file mode 100644
index 000000000..ddc76f37c
--- /dev/null
+++ b/kth-smallest-element-in-a-bst/jdalma.kt
@@ -0,0 +1,98 @@
+package leetcode_study
+
+import io.kotest.matchers.equals.shouldBeEqual
+import org.junit.jupiter.api.Test
+import java.util.PriorityQueue
+
+class `kth-smallest-element-in-a-bst` {
+
+    fun kthSmallest(root: TreeNode?, k: Int): Int {
+        return inorderTraversal(root, k)
+    }
+
+    // 1. 재귀 호출로 모든 트리를 조회 후 정렬
+    // 시간복잡도: O(n * log(n)), 공간복잡도: O(n)
+    private fun recursionAndSort(root: TreeNode?, k: Int) = mutableSetOf<Int>().apply {
+        dfs(root, this)
+    }.sorted()[k - 1]
+
+    private fun dfs(node: TreeNode?, set: MutableSet<Int>) {
+        if (node == null) return
+
+        set.add(node.`val`)
+        dfs(node.left, set)
+        dfs(node.right, set)
+    }
+
+    // 2. 재귀 호출로 모든 트리의 값을 우선순위 큐에 삽입하고 작은 값으로 계속 큐를 갱신
+    // 시간복잡도: O(n * log(k)), 공간복잡도: O(n + k)
+    // 트리 순회 : O(n), 우선순위 큐 삽입: O(log k)
+    private fun usingPriorityQueue(node: TreeNode?, k: Int): Int {
+        fun dfs(node: TreeNode, k: Int, pq: PriorityQueue<Int>) {
+            pq.offer(node.`val`)
+
+            if (pq.size > k) {
+                pq.poll()
+            }
+            if (node.left != null) {
+                dfs(node.left!!, k, pq)
+            }
+            if (node.right != null) {
+                dfs(node.right!!, k, pq)
+            }
+        }
+
+        val pq = PriorityQueue { v1: Int, v2: Int -> v2 - v1 }
+        dfs(node!!, k, pq)
+        return pq.first()
+    }
+
+    // 3. 문제의 전제가 이진탐색트리이기에, 중위순회로 탐색하여 값을 누적하면 오름차순의 값이 된다.
+    // 시간복잡도: O(n), 공간복잡도: O(n)
+    private fun inorderTraversal(node: TreeNode?, k: Int): Int {
+        fun dfs(node: TreeNode, k: Int, list: MutableList<Int>) {
+            if (node.left != null) {
+                dfs(node.left!!, k, list)
+            }
+            list.add(node.`val`)
+            if (node.right != null) {
+                dfs(node.right!!, k, list)
+            }
+        }
+
+        return mutableListOf<Int>().apply {
+            dfs(node!!, k, this)
+        }[k - 1]
+    }
+
+    @Test
+    fun `루트와 정수 k가 주어지면 트리에 있는 모든 노드의 값 중 가장 작은 값을 반환한다`() {
+        kthSmallest(TreeNode.of(listOf(0, 3,1,4,null,2)), 1) shouldBeEqual 1
+        kthSmallest(TreeNode.of(listOf(0, 5,3,6,2,4,null,null,1)), 3) shouldBeEqual 3
+    }
+}
+
+class TreeNode(var `val`: Int) {
+    var left: TreeNode? = null
+    var right: TreeNode? = null
+
+    companion object {
+        fun of(numbers: List<Int?>): TreeNode? {
+            fun setChild(node: TreeNode?, nums: List<Int?>, index: Int): TreeNode? {
+                if (node == null) return null
+                val (leftIndex, rightIndex) = index * 2 to index * 2 + 1
+
+                if (leftIndex < nums.size && nums[leftIndex] != null) {
+                    node.left = TreeNode(nums[leftIndex]!!)
+                    setChild(node.left, nums, leftIndex)
+                }
+                if (rightIndex < nums.size && nums[rightIndex] != null) {
+                    node.right = TreeNode(nums[rightIndex]!!)
+                    setChild(node.right, nums, rightIndex)
+                }
+                return node
+            }
+            return setChild(TreeNode(numbers[1]!!), numbers, 1)
+        }
+    }
+}
diff --git a/number-of-1-bits/jdalma.kt b/number-of-1-bits/jdalma.kt
new file mode 100644
index 000000000..eea9462fb
--- /dev/null
+++ b/number-of-1-bits/jdalma.kt
@@ -0,0 +1,46 @@
+package leetcode_study
+
+import io.kotest.matchers.equals.shouldBeEqual
+import org.junit.jupiter.api.Test
+
+class `number-of-1-bits` {
+
+    fun hammingWeight(n: Int): Int {
+        return bitOperation(n)
+    }
+
+    // 1. 직접 2진수를 구함
+    // 시간복잡도: O(log n), 공간복잡도: O(1)
+    private fun binary(n: Int): Int {
+        var calc = n
+        var count = 0
+        while(calc > 0) {
+            if (calc % 2 != 0) {
+                count++
+            }
+            calc /= 2
+        }
+        return count
+    }
+
+    // 2. 비트 논리 연산자 사용
+    // 시간복잡도: O(log n), 공간복잡도: O(1)
+    private fun bitOperation(n: Int): Int {
+        var calc = n
+        var count = 0
+        while (calc > 0) {
+            if (calc and 1 == 1) {
+                count ++
+            }
+            calc = calc shr 1
+        }
+        return count
+    }
+
+    @Test
+    fun `이진수에서_0이_아닌_성분의_개수를 반환한다`() {
+        hammingWeight(11) shouldBeEqual 3
+        hammingWeight(128) shouldBeEqual 1
+        hammingWeight(2147483645) shouldBeEqual 30
+    }
+}
diff --git a/palindromic-substrings/jdalma.kt b/palindromic-substrings/jdalma.kt
new file mode 100644
index 000000000..6aa5200ff
--- /dev/null
+++ b/palindromic-substrings/jdalma.kt
@@ -0,0 +1,90 @@
+package leetcode_study
+
+import io.kotest.matchers.equals.shouldBeEqual
+import org.junit.jupiter.api.Test
+
+class `palindromic-substrings` {
+
+    fun countSubstrings(s: String): Int {
+        return dp(s)
+    }
+
+    // 1. 브루트포스 3중 반복문
+    // 시간복잡도: O(n^3), 공간복잡도: O(1)
+    private fun bruteForce(s: String): Int {
+
+        fun isPalindrome(text: String, left: Int, right: Int): Boolean {
+            var (start, end) = left to right
+            while (start < end) {
+                if (text[start] != text[end]) {
+                    return false
+                }
+                start++
+                end--
+            }
+            return true
+        }
+
+        var count = 0
+        s.indices.forEachIndexed { left, _ ->
+            (left until s.length).forEach { right ->
+                if (isPalindrome(s, left, right)) {
+                    count++
+                }
+            }
+        }
+        return count
+    }
+
+    // 2. 문자열의 길이가 홀수,짝수를 감안하여 특정 지점부터 두 개의 포인터 left , right를 비교한다.
+    // 시간복잡도: O(n^2), 공간복잡도: O(1)
+    private fun twoPointer(s: String): Int {
+
+        fun palindromeCount(text: String, left: Int, right: Int): Int {
+            var count = 0
+            var (l , r) = left to right
+            while (l >= 0 && r < text.length && text[l--] == text[r++]) {
+                count++
+            }
+            return count
+        }
+
+        var result = 0
+        s.indices.forEachIndexed { index, e ->
+            val even = palindromeCount(s, index, index + 1)
+            val odd = palindromeCount(s, index - 1, index + 1)
+            result += even + odd + 1
+        }
+
+        return result
+    }
+
+    // 3. DP : 이전에 비교했던 결과를 기억하여 (현재 주어진 start, end  비교) && (이전 start, end 비교)
+    // 시간복잡도: O(n^2), 공간복잡도: O(n^2)
+    private fun dp(s: String): Int {
+        val len = s.length
+        val dp = Array(len) { BooleanArray(len) { false } }
+
+        for (end in 0 until len) {
+            for (start in end downTo 0) {
+                if (start == end) {
+                    dp[start][end] = true
+                } else if (start + 1 == end) {
+                    dp[start][end] = s[start] == s[end]
+                } else {
+                    dp[start][end] = s[start] == s[end] && dp[start + 1][end - 1]
+                }
+            }
+        }
+
+        return dp.sumOf { row -> row.count { it } }
+    }
+
+    @Test
+    fun `주어진 문자열의 회문인 부분 문자열의 개수를 반환한다`() {
+        countSubstrings("abc") shouldBeEqual 3
+        countSubstrings("aaa") shouldBeEqual 6
+        countSubstrings("ababa") shouldBeEqual 9
+        countSubstrings("abcda") shouldBeEqual 5
+    }
+}
diff --git a/top-k-frequent-elements/jdalma.kt b/top-k-frequent-elements/jdalma.kt
new file mode 100644
index 000000000..4f1e05d49
--- /dev/null
+++ b/top-k-frequent-elements/jdalma.kt
@@ -0,0 +1,82 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+import java.util.PriorityQueue
+
+class `top-k-frequent-elements` {
+
+    fun topKFrequent(nums: IntArray, k: Int): IntArray {
+        return third(nums, k)
+    }
+
+    // 1. Map 정렬
+    // 시간복잡도: O(n * log(n)), 공간복잡도: O(n)
+    private fun first(nums: IntArray, k: Int): IntArray {
+        val map = mutableMapOf<Int, Int>()
+
+        nums.forEach {
+            map.compute(it) { _, oldValue ->
+                if (oldValue == null) 1
+                else oldValue + 1
+            }
+        }
+
+        return map.entries.sortedByDescending { it.value }
+            .map { it.key }
+            .slice(0 until k)
+            .toIntArray()
+    }
+
+    // 2. 우선순위 큐
+    // 시간복잡도: O(n * log(k)), 공간복잡도: O(n + k)
+    private fun second(nums: IntArray, k: Int): IntArray {
+        val map = mutableMapOf<Int, Int>()
+
+        nums.forEach { map.put(it, map.getOrDefault(it, 0) + 1) }
+
+        val heap: PriorityQueue<Map.Entry<Int, Int>> = PriorityQueue<Map.Entry<Int, Int>> {
+            v1, v2 -> v2.value.compareTo(v1.value)
+        }.apply {
+            this.addAll(map.entries)
+        }
+
+        return (0 until k).map { heap.poll().key }.toIntArray()
+    }
+
+    // 3. 이차원배열로 빈번도 저장
+    // 시간복잡도: O(n), 공간복잡도: O(n)
+    private fun third(nums: IntArray, k: Int): IntArray {
+        val map = mutableMapOf<Int, Int>()
+
+        nums.forEach { map.put(it, map.getOrDefault(it, 0) + 1) }
+
+        val freq = Array<MutableList<Int>>(nums.size + 1) { mutableListOf() }
+        map.entries.forEach {
+            val frequency = it.value
+            freq[frequency].add(it.key)
+        }
+
+        val result = IntArray(k)
+        var index = 0
+        (freq.size - 1 downTo 0).forEach { i ->
+            freq[i].forEach {
+                result[index++] = it
+                if (index == k) {
+                    return result
+                }
+            }
+        }
+
+        return IntArray(0)
+    }
+
+    @Test
+    fun `배열에서_가장_빈도가_높은_K개의_원소를_출력한다`() {
+        topKFrequent(intArrayOf(1,1,1,2,2,3), 2) shouldBe intArrayOf(1,2)
+        topKFrequent(intArrayOf(1,1,1,2,2,3,3,4), 3) shouldBe intArrayOf(1,2,3)
+        topKFrequent(intArrayOf(2,2,3,3,1,1,4), 3) shouldBe intArrayOf(2,3,1)
+        topKFrequent(intArrayOf(4,1,-1,2,-1,2,3), 2) shouldBe intArrayOf(-1,2)
+
+    }
+}