Combinations for Target Sum

mediumcodingAlgorithmsBacktracking

Overview

Given an array of distinct positive integers candidates and a target integer target, return the number of possible combinations that add up to target.

Each number in candidates may be used an unlimited number of times. Two combinations are different if the sequence of numbers differs (order matters).

Constraints

  • 1 <= candidates.length <= 200
  • 1 <= candidates[i] <= 1000
  • All elements of candidates are distinct.
  • 1 <= target <= 1000

Examples

combinationSum([1, 2, 3], 4);
// => 7
// The combinations are:
// (1,1,1,1), (1,1,2), (1,2,1), (1,3),
// (2,1,1), (2,2), (3,1)
combinationSum([9], 3);
// => 0  (no way to reach 3 with only 9s)

combinationSum([2, 3], 5);
// => 3  (2+3, 3+2, 2+2+... no, just: 2+3, 3+2, 2+2 won't work... => 2+3, 3+2 = 2 ... wait)
// Actually: (2,3), (3,2) = 2... let me recalc
// Actually for target 5 with [2,3]: 2+3=5, 3+2=5 => 2

Solution

Reveal solution
function combinationSum(candidates, target) {
  const dp = new Array(target + 1).fill(0);
  dp[0] = 1;
  for (let i = 1; i <= target; i++) {
    for (const c of candidates) {
      if (c <= i) dp[i] += dp[i - c];
    }
  }
  return dp[target];
}

This is a bottom-up DP where dp[i] counts the number of ordered combinations summing to i. For each amount, we try adding each candidate.

Resources

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