11var threeSum = function ( nums ) {
2+ // Create array a to store output array
23 let a = [ ]
4+ // Create variable j and k for indexing nums[j] and nums[k]
5+ // Create val1 and val2 for storing value of nums[j] and nums[k] respectively
36 let [ j , k , val1 , val2 ] = [ 0 , 0 , 0 , 0 ]
7+ // sort the input array ascendingly
48 nums . sort ( function ( a , b ) { return a - b } )
5- for ( let i = 0 ; i < nums . length - 2 ; i ++ ) {
9+ // run loop over nums from (0,nums.length-2)
10+ // we need i , j , k so last usable index of will be nums.length-2
11+ // storing length helps a little bit for optimization
12+ len = nums . length
13+ for ( let i = 0 ; i < len - 2 ; i ++ ) {
14+ // nums[i] must be negative or 0 to get nums[i] + nums[j] + nums[k] == 0
615 if ( nums [ i ] > 0 ) { continue }
16+ // we will cover all possible triplet for nums[i] so we dont need to rerun for same value
717 if ( nums [ i ] == nums [ i - 1 ] ) { continue }
18+ // j will be the start pointer and k will be end pointer for our window
819 j = 1 + i
9- k = nums . length - 1
20+ k = len - 1
21+ // we will run this until the whole window is convered
1022 while ( j < k ) {
1123 if ( nums [ i ] + nums [ j ] + nums [ k ] == 0 ) {
24+ // if this condition is met we add it to answer array
1225 a . push ( [ nums [ i ] , nums [ j ] , nums [ k ] ] )
26+ // we store value of nums[j] in val1
1327 val1 = nums [ j ] ;
28+ // we skip all index with same value as nums[j] this will help up shorten the window
1429 while ( j < k && nums [ j ] == val1 ) j ++ ;
1530 val2 = nums [ k ] ;
31+ // we skip all index with same value as nums[k] this will help up shorten the window
1632 while ( j < k && nums [ k ] == val2 ) k -- ;
17- } else if ( nums [ i ] + nums [ j ] + nums [ k ] < 0 ) j ++ ;
33+ // if total is < 0 we try to get a greater value for nums[j] (array is sorted ascending)
34+ } else if ( nums [ i ] + nums [ j ] + nums [ k ] < 0 ) j ++ ;
35+ // if total is > 0 we try to get a smaller value for nums[k] (array is sorted ascending)
1836 else if ( nums [ i ] + nums [ j ] + nums [ k ] > 0 ) k -- ;
37+ // thus when this loop is done we will have all possible j and k for given nums[i]
1938 }
2039 }
40+ // return the output array
2141 return a
22- } ;
42+ } ;
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