### Problem

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number of times.

Note: All numbers (including target) will be positive integers. Elements in a combination (a1, a2, ... , ak) must be in non-descending order. (ie, a1 <= a2 <= ... <= ak). The solution set must not contain duplicate combinations. For example, given candidate set 2,3,6,7 and target 7,

A solution set is:

[7]

[2, 2, 3]

### JavaScript Code

function combinationSum(candidates, target) { var result = []; if(candidates == null || candidates.length == 0) return result; var current = []; candidates.sort(); combinationSumHelper(candidates, target, 0, current, result); return result; } function combinationSumHelper(candidates, target, j, curr, result){ if(target == 0){ var temp = curr.slice(); result.push(temp); return; } for(var i=j; i<candidates.length; i++){ if(target < candidates[i]) return; curr.push(candidates[i]); combinationSumHelper(candidates, target - candidates[i], i, curr, result); curr.pop(); } }

### With Unique Combinations

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used ONCE in the combination.

Note:

1) All numbers (including target) will be positive integers.

2) Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

3) The solution set must not contain duplicate combinations.

### JavaScript Code

var result = []; function combinationSum(num, target) { result = []; if(num == null || num.length == 0) return result; num.sort(); var temp = []; getCombination(num, 0, target, temp, result); } function getCombination(num, start, target, temp){ if(target == 0){ var t = JSON.parse(JSON.stringify(temp)); if(indexOf(result, t, arraysIdentical) == -1) { result.push(t); } return; } for(var i=start; i<num.length; i++){ if(target < num[i]) continue; temp.push(num[i]); getCombination(num, i+1, target-num[i], temp, result); temp.pop(); } } function arraysIdentical(arr1, arr2) { var i = arr1.length; if (i !== arr2.length) { return false; } while (i--) { if (arr1[i] !== arr2[i]) { return false; } } return true; } function indexOf(arr, val, comparer) { for (var i = 0, len = arr.length; i < len; ++i) { if ( i in arr && comparer(arr[i], val) ) { return i; } } return -1; }

Excellent mind bending code :-)

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