diff --git a/longest-substring-without-repeating-characters/jdalma.kt b/longest-substring-without-repeating-characters/jdalma.kt
new file mode 100644
index 000000000..4a2837e96
--- /dev/null
+++ b/longest-substring-without-repeating-characters/jdalma.kt
@@ -0,0 +1,46 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+import kotlin.math.max
+
+class `longest-substring-without-repeating-characters` {
+
+    fun lengthOfLongestSubstring(s: String): Int {
+        if (s.length <= 1) return s.length
+        return usingSet(s)
+    }
+
+    /**
+     * 1. 사용한 문자를 Set에 저장하여 확인하고, 중복된다면 해당 문자의 위치까지 모든 문자를 제거한다.
+     * TC: O(n), SC: O(n)
+     */
+    private fun usingSet(s: String): Int {
+        var left = 0
+        val used = mutableSetOf<Char>()
+        var maxLength = 0
+
+        for (right in s.indices) {
+            if (!used.contains(s[right])) {
+                maxLength = max(right - left + 1, maxLength)
+                used.add(s[right])
+            } else {
+                while (used.contains(s[right])) {
+                    used.remove(s[left])
+                    left++
+                }
+                used.add(s[right])
+            }
+        }
+
+        return maxLength
+    }
+
+    @Test
+    fun `입력받은 문자열의 반복되는 문자가 없는 가장 긴 부분 문자열의 길이를 반환한다`() {
+        lengthOfLongestSubstring("ababc") shouldBe 3
+        lengthOfLongestSubstring("bbbbb") shouldBe 1
+        lengthOfLongestSubstring("abcabcbb") shouldBe 3
+        lengthOfLongestSubstring("pwwkew") shouldBe 3
+    }
+}
diff --git a/number-of-islands/jdalma.kt b/number-of-islands/jdalma.kt
new file mode 100644
index 000000000..48f95f81e
--- /dev/null
+++ b/number-of-islands/jdalma.kt
@@ -0,0 +1,83 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+
+class `number-of-islands` {
+
+    private data class Position(
+        val x: Int,
+        val y: Int
+    ) {
+
+        operator fun plus(other: Position) = Position(this.x + other.x, this.y + other.y)
+
+        fun isNotOutOfIndexed(board: Array<CharArray>) =
+            this.x < board.size && this.x >= 0 && this.y < board[0].size &&  this.y >= 0
+
+        companion object {
+            val MOVES: List<Position> = listOf(
+                Position(-1, 0),
+                Position(0, 1),
+                Position(1, 0),
+                Position(0, -1),
+            )
+        }
+    }
+
+    /**
+     * BFS를 사용한 탐색
+     * TC: O(n * m), SC: O(n * m)
+     */
+    fun numIslands(grid: Array<CharArray>): Int {
+        val (row, col) = grid.size to grid.first().size
+        val visited = Array(row) { BooleanArray(col) }
+        var result = 0
+
+        for (x in 0 until row) {
+            for (y in 0 until col) {
+                if (!visited[x][y] && grid[x][y] == '1') {
+                    visited[x][y] = true
+                    bfs(x, y, grid, visited)
+                    result++
+                }
+            }
+        }
+        return result
+    }
+
+    private fun bfs(x: Int, y: Int, grid: Array<CharArray>, visited: Array<BooleanArray>) {
+        val queue = ArrayDeque<Position>().apply {
+            this.add(Position(x, y))
+        }
+
+        while (queue.isNotEmpty()) {
+            val now = queue.removeFirst()
+            for (move in Position.MOVES) {
+                val moved = now + move
+                if (moved.isNotOutOfIndexed(grid) && grid[moved.x][moved.y] == '1' && !visited[moved.x][moved.y]) {
+                    visited[moved.x][moved.y] = true
+                    queue.add(moved)
+                }
+            }
+        }
+    }
+
+
+    @Test
+    fun `문자 이차원배열을 입력받으면 1로 이루어진 영역의 수를 반환한다`() {
+        numIslands(arrayOf(
+            charArrayOf('1','1','1','1','0'),
+            charArrayOf('1','1','0','1','0'),
+            charArrayOf('1','1','0','0','0'),
+            charArrayOf('0','0','0','0','0')
+        )) shouldBe 1
+
+        numIslands(arrayOf(
+            charArrayOf('1','1','0','0','0'),
+            charArrayOf('1','1','0','0','0'),
+            charArrayOf('0','0','1','0','0'),
+            charArrayOf('0','0','0','1','1')
+        )) shouldBe 3
+    }
+}
diff --git a/reverse-linked-list/jdalma.kt b/reverse-linked-list/jdalma.kt
new file mode 100644
index 000000000..b61e3aad1
--- /dev/null
+++ b/reverse-linked-list/jdalma.kt
@@ -0,0 +1,36 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+
+class `reverse-linked-list` {
+
+    /**
+     * TC : O(n), SC: O(1)
+     */
+    fun reverseList(head: ListNode?): ListNode? {
+        var prev: ListNode? = null
+        var next: ListNode? = null
+        var curr: ListNode? = head
+
+        while (curr != null) {
+            next = curr.next
+            curr.next = prev
+            prev = curr
+            curr = next
+        }
+
+        return prev
+    }
+
+    @Test
+    fun `입력 받은 단일 리스트를 반전하고 반전한 목록을 반환한다`() {
+        reverseList(ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5))))))
+            shouldBe(ListNode(5, ListNode(4, ListNode(3, ListNode(2, ListNode(1))))))
+    }
+}
+
+class ListNode(
+    var `val`: Int,
+    var next: ListNode? = null
+)
diff --git a/set-matrix-zeroes/jdalma.kt b/set-matrix-zeroes/jdalma.kt
new file mode 100644
index 000000000..477969ccc
--- /dev/null
+++ b/set-matrix-zeroes/jdalma.kt
@@ -0,0 +1,111 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+import kotlin.math.max
+
+class `set-matrix-zeroes` {
+
+    private data class Position(
+        val x: Int,
+        val y: Int
+    )
+
+    fun setZeroes(matrix: Array<IntArray>): Unit {
+        usingFlag(matrix)
+    }
+
+    /**
+     * 0으로 변경해야 할 열과 행을 Set에 담아 처리한다.
+     * TC: O(n * m) SC: O(n + m)
+     */
+    private fun usingSet(matrix: Array<IntArray>) {
+        val zeroRows = mutableSetOf<Int>()
+        val zeroCols = mutableSetOf<Int>()
+        for (i in matrix.indices) {
+            for (j in matrix.first().indices) {
+                if (matrix[i][j] == 0) {
+                    zeroRows.add(i)
+                    zeroCols.add(j)
+                }
+            }
+        }
+
+        for (row in zeroRows) {
+            for (col in matrix.first().indices) {
+                matrix[row][col] = 0
+            }
+        }
+
+        for (col in zeroCols) {
+            for (row in matrix.indices) {
+                matrix[row][col] = 0
+            }
+        }
+    }
+
+    /**
+     * 0으로 변경해야 할 열과 행을 matrix 0번째 행과 0번째 열 그리고 두 개의 flag로 처리하여 공간복잡도를 개선한다.
+     * TC: O(n * m) SC: O(1)
+     */
+    private fun usingFlag(matrix: Array<IntArray>) {
+        var (rowFlag, colFlag) = false to false
+        for (i in matrix.indices) {
+            for (j in matrix.first().indices) {
+                if (matrix[i][j] == 0) {
+                    if (i == 0) rowFlag = true
+                    if (j == 0) colFlag = true
+                    matrix[0][j] = 0
+                    matrix[i][0] = 0
+                }
+            }
+        }
+
+        for (i in 1 until matrix.size) {
+            for (j in 1 until matrix.first().size) {
+                if (matrix[i][0] == 0 || matrix[0][j] == 0) {
+                    matrix[i][j] = 0
+                }
+            }
+        }
+
+        if (rowFlag) {
+            for (i in matrix.first().indices) {
+                matrix[0][i] = 0
+            }
+        }
+
+        if (colFlag) {
+            for (element in matrix) {
+                element[0] = 0
+            }
+        }
+    }
+
+    @Test
+    fun `원소가 0이라면 해당 행과 열을 모두 0으로 수정한다`() {
+        val actual1 = arrayOf(
+            intArrayOf(1,1,1),
+            intArrayOf(1,0,1),
+            intArrayOf(1,1,1)
+        )
+        setZeroes(actual1)
+        actual1 shouldBe arrayOf(
+            intArrayOf(1,0,1),
+            intArrayOf(0,0,0),
+            intArrayOf(1,0,1)
+        )
+
+        val actual2 = arrayOf(
+            intArrayOf(0,1,2,0),
+            intArrayOf(3,4,5,2),
+            intArrayOf(1,3,1,5)
+        )
+        setZeroes(actual2)
+        actual2 shouldBe arrayOf(
+            intArrayOf(0,0,0,0),
+            intArrayOf(0,4,5,0),
+            intArrayOf(0,3,1,0)
+        )
+    }
+}
diff --git a/unique-paths/jdalma.kt b/unique-paths/jdalma.kt
new file mode 100644
index 000000000..53ecda265
--- /dev/null
+++ b/unique-paths/jdalma.kt
@@ -0,0 +1,58 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+
+/**
+ * 0,0 에서 아래 또는 오른쪽으로만 이동 가능하다.
+ */
+class `unique-paths` {
+
+    fun uniquePaths(row: Int, col: Int): Int {
+        return if (row <= 1 || col <= 1) 1
+        else usingArray(row, col)
+    }
+
+    /**
+     * TC: O(n * m), SC: O(n * m)
+     */
+    private fun usingGrid(row: Int, col: Int): Int {
+        val grid = Array(row) { IntArray(col) }
+        (0 until row).forEach { grid[it][0] = 1 }
+        (0 until col).forEach { grid[0][it] = 1 }
+
+        for (i in 1 until row) {
+            for (j in 1 until col) {
+                grid[i][j] = grid[i - 1][j] + grid[i][j - 1]
+            }
+        }
+
+        return grid[row - 1][col - 1]
+    }
+
+    /**
+     * 이전 라인의 배열만 기억하여도 되므로 공간 복잡도를 아래와 같이 줄일 수 있다.
+     * TC: O(n * m), SC: O(m)
+     */
+    private fun usingArray(row: Int, col: Int): Int {
+        var dp = IntArray(col)
+
+        for (i in 0 until row) {
+            val tmp = IntArray(col)
+            for (j in 0 until col) {
+                if (i == 0 && j == 0) tmp[j] = 1
+                else if (j > 0) tmp[j] = dp[j] + tmp[j - 1]
+                else tmp[j] = dp[j]
+            }
+            dp = tmp
+        }
+
+        return dp.last()
+    }
+
+    @Test
+    fun `왼쪽 상단 모서리에서 오른쪽 상단 모서리로 도달할 수 있는 고유 경로의 수를 반환한다`() {
+        uniquePaths(3, 7) shouldBe 28
+        uniquePaths(3, 2) shouldBe 3
+    }
+}