#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/math-fundamentals/combinations/combination-sum-implementation-3

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░████████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/math-fundamentals/combinations/combination-sum-implementation-3

Repovive

Combination Sum - Implementation - Combinations | Math Fundamentals | Repovive

Repovive.

Math Fundamentals18 sections · 842 units13h total

Open in Course

Combination Sum - Implementation

The pseudocode

Here is the full solution:


function combinationSum(candidates, target)
    result := []
    backtrack(candidates, 0, [], 0, target, result)
    return result

function backtrack(candidates, start, current, sum, target, result)
    if sum = target then
        add copy of current to result
        return
    if sum > target then
        return
    for i from start to length of candidates - 1
        add candidates[i] to current
        backtrack(candidates, i, current, sum + candidates[i], target, result)
        remove last element from current

Time: $O(2^t)$ where $t$ is the target value (worst case when smallest candidate is $1$ ). Space: $O(t)$ for recursion depth.