DSA Visualizations | Hello Interview

Combination Sum

DESCRIPTION (credit Leetcode.com)

Given an array of distinct integers candidates and a target integer target, generate all unique combinations of candidates which sum to target. The combinations may be returned in any order, and the same number may be chosen from candidates an unlimited number of times.

EXAMPLES

Input:


candidates = [2,3,6,7]
target = 7

Output:

[[2,2,3],[7]]

Explanation:

2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations.

Solution


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result

combination sum

0 / 56

Explanation

This solution uses backtracking to find all the combinations of numbers that sum up to the target. The backtracking function is a recursive function that uses depth-first search to explore all possible combinations of candidates that sum up to the target. Each call to backtracking takes in parameters start, combo, and current_target. The start parameter is the index of the current candidate, combo is the current combination of numbers, and the current_target parameter is the target we are trying to hit (equal to target - the sum of combo).

The first step is to sort candidates, which makes it easy to tell when the current search combination exceeds the target, and can be pruned.


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result

combination sum

0 / 1

Next, we kick off the search. Each call to the backtracking function first checks if the current_target is equal to 0. If so, that means the current value for combo is a valid solution, so we add it to the result. If not, the function iterates through the candidates starting from the start index. For each candidate, we add it to the current combination and recursively call the backtracking function with the updated current_target. Since we can use the same candidate multiple times, we pass the start index as the current index to the next call.


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result

sort candidates

0 / 10

Recursively exploring all combinations

If at any point the current_target becomes less than 0, we know that the current combo is not a valid solution, so we backtrack to the previous call.

For the backtracking step, we remove the last number from the current combination and try the next candidate. This process continues until we have tried all the candidates.


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return

i = 0

0 / 2

Backtracking to the previous call. Notice how we pop from combo when we backtrack.

Next: Overview

Your account is free and you can post anonymously if you choose.

Sort By

Loading comments...

Back to Home

Eliminating Pairs

Overview

Container With Most Water

3-Sum

Triangle Numbers

Pointers as Regions

Move Zeroes

Sort Colors

Trapping Rain Water

Variable-Length

Overview

Longest Substring Without Repeating Characters

Longest Repeating Character Replacement

Fixed-Length

Overview

Maximum Sum of Subarrays of Size K

Max Points You Can Obtain From Cards

Max Sum of Distinct Subarrays Length k

Overview

Can Attend Meetings

Insert Interval

Non-Overlapping Intervals

Merge Intervals

Employee Free Time

Overview

Valid Parentheses

Decode String

Longest Valid Parentheses

Monotonic Stack

Overview

Daily Temperatures

Largest Rectangle in Histogram

Overview

Linked List Cycle

Palindrome Linked List

Remove Nth Node From End of List

Reorder List

Swap Nodes in Pairs

Overview

Apple Harvest (Koko Eating Bananas)

Search in Rotated Sorted Array

Overview

Kth Largest Element in an Array

K Closest Points to Origin

Find K Closest Elements

Merge K Sorted Lists

Introduction

Binary Trees

Overview

Level Order Sum

Rightmost Node

Zigzag Level Order

Maximum Width of Binary Tree

Graphs

Overview

Minimum Knight Moves

Rotting Oranges

01-Matrix

Bus Routes

Overview

Word Search

Solution Space Trees

Subsets

Generate Parentheses

Combination Sum

Topological Sort

Overview

Course Schedule

Course Schedule II

Fundamentals

Solving a Question with Dynamic Programming

Counting Bits

Decode Ways

Maximal Square

Unique Paths

Longest Increasing Subsequence

Word Break

Maximum Profit in Job Scheduling

Overview

Best Time to Buy and Sell Stock

Gas Station

Jump Game

Overview

Implement Trie Methods

Prefix Matching

Overview

Count Vowels in Substrings

Subarray Sum Equals K

Spiral Matrix

Rotate Image

Set Matrix Zeroes

Generate Parentheses

Graphs: Overview

Combination Sum

DESCRIPTION (credit Leetcode.com)

EXAMPLES

Input:


candidates = [2,3,6,7]
target = 7

Output:

[[2,2,3],[7]]

Explanation:

2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations.

Solution


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result

combination sum

0 / 56

Explanation

The first step is to sort candidates, which makes it easy to tell when the current search combination exceeds the target, and can be pruned.


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result

combination sum

0 / 1


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result

sort candidates

0 / 10

Recursively exploring all combinations

If at any point the current_target becomes less than 0, we know that the current combo is not a valid solution, so we backtrack to the previous call.

For the backtracking step, we remove the last number from the current combination and try the next candidate. This process continues until we have tried all the candidates.


def combinationSum(candidates, target):
  def backtrack(start, combo, current_target):
    if current_target == 0:
      result.append(list(combo))
      return
    for i in range(start, len(candidates)):
      curr = candidates[i]
      if candidates[i] > current_target:
        return
      combo.append(curr)
      backtrack(i, combo, current_target - curr)
      combo.pop()
    return

  candidates.sort()
  result = []
  backtrack(0, [], target)
  return result


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return


def backtrack(start, combo, current_target):
  if current_target == 0:
    result.append(list(combo))
    return
  for i in range(start, len(candidates)):
    curr = candidates[i]
    if curr > current_target:
      return
    combo.append(curr)
    backtrack(i, combo, current_target - curr)
    combo.pop()
  return

i = 0

0 / 2

Backtracking to the previous call. Notice how we pop from combo when we backtrack.

Next: Overview

Your account is free and you can post anonymously if you choose.

Sort By

Loading comments...

BlogBehavioral Interview Examples Interviewing for Experienced SWEs All Blog Posts

Compare UsCompare to Interviewing.io

LinksFAQ Schedule Mock Interviews Pricing Become a Coach Our Coaches Learn System Design Learn DSA Guided Practice Hello Interview Premium

LegalTerms and Conditions Privacy Policy

Contactsupport@hellointerview.com7511 Greenwood Ave North
Unit #4238 Seattle, WA 98103