Add combination-sum solution in TypeScript

Author: Jeehay28

combination-sum/Jeehay28.ts

@@ -0,0 +1,82 @@
+// Approach 2: Dynamic Programming
+// ✅ Time Complexity: O(N * T * K)
+// ✅ Space Complexity: O(T * K)
+// N = Number of candidates, T = target, K = average number of combination for each dp[i]
+
+function combinationSum(candidates: number[], target: number): number[][] {
+  // candidates = [2, 3]
+  // target = 5
+  // dp = [
+  //     [[]],         // dp[0]
+  //     [],           // dp[1]
+  //     [[2]],        // dp[2]
+  //     [],           // dp[3]
+  //     [[2, 2]],     // dp[4]
+  //     []            // dp[5]
+  // ];
+
+  // dp = [
+  //     [[]],         // dp[0]
+  //     [],           // dp[1]
+  //     [[2]],        // dp[2]
+  //     [[3]],        // dp[3]
+  //     [[2, 2]],     // dp[4]
+  //     [[2, 3]]      // dp[5]
+  // ];
+
+  const dp: number[][][] = Array.from({ length: target + 1 }, () => []);
+  // each element in dp is an independent array, and modifying one will not affect others.
+  dp[0] = [[]];
+
+  for (const candidate of candidates) {
+    for (let num = candidate; num <= target; num++) {
+      for (const combination of dp[num - candidate]) {
+        dp[num].push([...combination, candidate]);
+      }
+    }
+  }
+
+  return dp[target];
+}
+
+// Approach 1: DFS + Backtracking (Recursive)
+// ✅ Time Complexity: O(N^(T / min))
+// ✅ Space Complexity: O(K + target / min)
+// If target = 7 and smallest number is 2, recursion can go up to  7 / 2 = levels deep
+// N = number of candidates, T = target value, K = total number of valid combination found
+
+// function combinationSum(candidates: number[], target: number): number[][] {
+//   // input:
+//   // 1) an array of distinct integers
+//   // 2) a target integer
+
+//   // output:
+//   // a list of all unique combinations of candinates where the chosen numbers sum to target(in any order)
+//   // duplicated numbers allowed
+
+//   let result: number[][] = [];
+//   let nums: number[] = [];
+
+//   const dfs = (start: number, total: number) => {
+//     if (total > target) {
+//       return;
+//     }
+
+//     if (total === target) {
+//       result.push([...nums]);
+//       return;
+//     }
+
+//     for (let i = start; i < candidates.length; i++) {
+//       if (total + nums[i] <= target) {
+//         nums.push(candidates[i]);
+//         dfs(i, total + candidates[i]);
+//         nums.pop(); // backtrack
+//       }
+//     }
+//   };
+
+//   dfs(0, 0);
+
+//   return result;
+// }