Explain the Problem
Find all possible combinations of k
numbers that add up to a number n
, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
All numbers will be positive integers, and no duplicate combinations are allowed.
Example:
Input: k = 3
, n = 7
Output:
[
[1, 2, 4]
]
1) Explain the problem
Given two integers k
and n
, return all possible combinations of k
numbers that add up to a number n
, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.
2) Short easy to remember solution/approach
Use backtracking to try different combinations.
Ensure the combination length is
k
and the sum isn
.
3) Solution in JavaScript with code commenting
function combinationSum3(k, n) {
const result = [];
function backtrack(start, currentCombination, currentSum) {
if (currentCombination.length === k && currentSum === n) {
result.push([...currentCombination]);
return;
}
if (currentCombination.length > k || currentSum > n) {
return;
}
for (let i = start; i <= 9; i++) {
currentCombination.push(i);
backtrack(i + 1, currentCombination, currentSum + i);
currentCombination.pop();
}
}
backtrack(1, [], 0);
return result;
}
// Example usage:
const k = 3;
const n = 7;
console.log(combinationSum3(k, n)); // Output: [[1, 2, 4]]
4) Explanation the solution in an easy-to-understand way with real-life example
Imagine you have a list of numbers from 1 to 9. You want to find all possible ways to pick k
numbers from this list so that their sum is exactly n
.
For example, if you want to pick 3 numbers (k = 3) that add up to 7 (n = 7), one possible way is to pick 1, 2, and 4 because 1 + 2 + 4 = 7.
5) Code explanation in pointers
Initialize result array: This array will store all valid combinations.
Backtracking function: This function will try different combinations of numbers.
Base case: If the current combination has
k
numbers and their sum isn
, add it to the result.Early stopping: If the combination length exceeds
k
or the sum exceedsn
, stop further exploration.Loop through numbers: From the current number to 9, try each number in the combination.
Recursive call: Add the current number to the combination and recurse to the next number.
Backtrack: Remove the last number from the combination to try the next possibility.
Return result: After exploring all possibilities, return the result array.
6) Complexities
Time complexity: (O(2^9)) because there are 9 numbers and each number can either be included or not in the combination.
Space complexity: (O(k)) for the recursion stack and combination array.