What if you could instantly find every way to combine numbers to hit your exact target without missing any?
Why Generate All Combinations Sum K in DSA Typescript?
Imagine you have a box of different colored beads, and you want to find all the ways to pick some beads so that their values sum to exactly 5. Doing this by guessing or writing down every possible group by hand would take forever!
Trying to list all groups manually is slow and confusing. You might miss some groups or repeat others. It's hard to keep track of which combinations you already tried, especially as the number of beads grows.
Using a method to generate all combinations that add up to a target number helps you find every possible group quickly and without mistakes. This method tries each bead, adds it if it fits, and moves on, backtracking when needed to explore all options.
let results = []; // Manually write all groups that sum to 5 results.push([1,4]); results.push([2,3]); // ... and so on, missing many combinations
function findCombinations(nums: number[], target: number) {
const results: number[][] = [];
function backtrack(start: number, path: number[], sum: number) {
if (sum === target) {
results.push([...path]);
return;
}
if (sum > target) return;
for (let i = start; i < nums.length; i++) {
path.push(nums[i]);
backtrack(i + 1, path, sum + nums[i]);
path.pop();
}
}
backtrack(0, [], 0);
return results;
}This lets you quickly find every possible group of numbers that add up to a target, unlocking solutions for puzzles, budgeting, and planning tasks.
Suppose you want to split a bill exactly using different coin values. Generating all combinations that sum to the bill amount helps you find all ways to pay without leftovers.
Manual listing is slow and error-prone.
Backtracking finds all combinations efficiently.
Useful for puzzles, budgeting, and planning.
Practice
Solution
Step 1: Understand the problem goal
The problem asks to find all unique sets of numbers whose sum equals K.Step 2: Clarify uniqueness and sum condition
Each combination must sum exactly to K without duplicates in the result.Final Answer:
All unique combinations of numbers that add up to K -> Option CQuick Check:
Combinations sum to K = All unique combinations of numbers that add up to K [OK]
- Confusing sum with count
- Including duplicate combinations
- Looking for single numbers only
Solution
Step 1: Analyze parameter types
The function should take an array of numbers and a target number.Step 2: Analyze return type
It should return an array of arrays of numbers representing combinations.Final Answer:
function generateCombinations(nums: number[], target: number): number[][] -> Option BQuick Check:
Input array and target number, output array of number arrays [OK]
- Using wrong parameter types
- Returning flat array instead of array of arrays
- Mixing string and number types
function combineSum(nums: number[], target: number): number[][] {
const result: number[][] = [];
function backtrack(start: number, path: number[], sum: number) {
if (sum === target) {
result.push([...path]);
return;
}
if (sum > target) return;
for (let i = start; i < nums.length; i++) {
path.push(nums[i]);
backtrack(i + 1, path, sum + nums[i]);
path.pop();
}
}
backtrack(0, [], 0);
return result;
}
console.log(combineSum([1, 2, 3], 3));Solution
Step 1: Trace backtracking calls
Starting from 0, combinations are built adding numbers without reuse: [1,2] sums to 3, [3] sums to 3.Step 2: Confirm valid combinations
Other combinations like [2] or [1,3] do not sum to 3 or are invalid due to index increment.Final Answer:
[[1, 2], [3]] -> Option AQuick Check:
Combinations summing to 3 = [[1, 2], [3]] [OK]
- Including combinations with reused elements
- Ignoring sum > target condition
- Confusing order of elements
function findCombos(nums: number[], target: number): number[][] {
const res: number[][] = [];
function dfs(i: number, path: number[], total: number) {
if (total === target) {
res.push(path);
return;
}
if (total > target || i === nums.length) return;
dfs(i + 1, path, total);
path.push(nums[i]);
dfs(i + 1, path, total + nums[i]);
path.pop();
}
dfs(0, [], 0);
return res;
}Solution
Step 1: Check how path is stored
The path array is pushed directly without copying, so later changes affect stored results.Step 2: Understand impact on results
Because path is referenced, all stored combinations end up empty or incorrect after backtracking.Final Answer:
The path array is pushed by reference causing incorrect results -> Option DQuick Check:
Push copy of path, not reference [OK]
- Pushing path without copying
- Not handling base cases properly
- Incorrect index increment
nums = [1, 2, 2, 3], target = 4
Solution
Step 1: Sort the input array
Sorting groups duplicates together, making it easier to skip them during recursion.Step 2: Skip duplicates in recursion
When iterating, skip elements equal to the previous to avoid duplicate combinations.Final Answer:
Sort the array and skip duplicates during recursion -> Option AQuick Check:
Sorting + skipping duplicates = unique combinations [OK]
- Not sorting before recursion
- Filtering duplicates after generation (inefficient)
- Ignoring duplicates causing repeated results
Ready to prove it?
Pass the test and this pattern moves unknown
