3Sum.cpp

// Source : https://oj.leetcode.com/problems/3sum/
// Author : Hao Chen
// Date   : 2014-07-22

/********************************************************************************** 
* 
* Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? 
* Find all unique triplets in the array which gives the sum of zero.
* 
* Note:
* 
* Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
* The solution set must not contain duplicate triplets.
* 
*     For example, given array S = {-1 0 1 2 -1 -4},
* 
*     A solution set is:
*     (-1, 0, 1)
*     (-1, -1, 2)
* 
*               
**********************************************************************************/

#include <stdio.h>
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;


/*
 *   Similar like "Two Number" problem, we can have the simlar solution.
 *
 *   Suppose the input array is S[0..n-1], 3SUM can be solved in O(n^2) time on average by 
 *   inserting each number S[i] into a hash table, and then for each index i and j,  
 *   checking whether the hash table contains the integer - (s[i]+s[j])
 *
 *   Alternatively, the algorithm below first sorts the input array and then tests all 
 *   possible pairs in a careful order that avoids the need to binary search for the pairs 
 *   in the sorted list, achieving worst-case O(n^n)
 *
 *   Solution:  Quadratic algorithm
 *   http://en.wikipedia.org/wiki/3SUM
 *
 */
vector<vector<int> > threeSum(vector<int> &num) {

    vector< vector<int> > result;
    if(num.size()==0 || num.size()==1 || num.size() == 2) return result;

    //sort the array, this is the key
    sort(num.begin(), num.end());

    int n = num.size();

    for (int i=0; i<n-2; i++) {
        //skip the duplication
        if (i>0 && num[i-1]==num[i]) continue;
        int a = num[i];
        int low = i+1;
        int high = n-1;
        while ( low < high ) {
            int b = num[low];
            int c = num[high];
            if (a+b+c == 0) {
                //got the soultion
                vector<int> v;
                v.push_back(a);
                v.push_back(b);
                v.push_back(c);
                result.push_back(v);
                // Continue search for all triplet combinations summing to zero.
                //skip the duplication
                while(low<n-1 && num[low]==num[low+1]) low++; 
                while(high>0 && num[high]==num[high-1]) high--; 
                low++;
                high--;
            } else if (a+b+c > 0) {
                //skip the duplication
                while(high>0 && num[high]==num[high-1]) high--;
                high--;
            } else{
                //skip the duplication
                while(low<n-1 && num[low]==num[low+1]) low++;
                low++;
            } 
        }
    }
    return result;
}

//using combination method could meet <<Time Limit Exceeded>> error
vector<vector<int> > combination(vector<int> &v, int k);
bool isSumZero(vector<int>& v);
int sum(vector<int>& v);

vector<vector<int> > threeSum2(vector<int> &num) {
    vector< vector<int> > result;
    vector< vector<int> > r = combination(num, 3);
    for (int i=0; i<r.size(); i++){
        if (isSumZero(r[i])){
            result.push_back(r[i]);
        }
    }
    return result;
}

bool isSumZero(vector<int>& v){
    return sum(v)==0;
}

int sum(vector<int>& v){
    int s=0;
    for(int i=0; i<v.size(); i++){
        s += v[i];
    }
    return s;
}

vector<vector<int> > combination(vector<int> &v, int k) {

    vector<vector<int> > result;
    vector<int> d;
    int n = v.size();
    for (int i=0; i<n; i++){
        d.push_back( (i<k) ? 1 : 0 );
    }

    //1) from the left, find the [1,0] pattern, change it to [0,1]
    //2) move all of the 1 before the pattern to the most left side
    //3) check all of 1 move to the right
    while(1){
        vector<int> tmp;
        for(int x=0; x<n; x++){
            if (d[x]) tmp.push_back(v[x]);
        }
        sort(tmp.begin(), tmp.end());
        result.push_back(tmp);
        //step 1), find [1,0] pattern
        int i;
        bool found = false;
        int ones =0;
        for(i=0; i<n-1; i++){

            if (d[i]==1 && d[i+1]==0){
                d[i]=0; d[i+1]=1;
                found = true;
                //step 2) move all of right 1 to the most left side
                for (int j=0; j<i; j++){
                    d[j]=( ones > 0 ) ? 1 : 0;
                    ones--;
                }
                break;
            }
            if (d[i]==1) ones++;
        }
        if (!found){
            break;
        }

    }
    return result;
}


void printMatrix(vector<vector<int> > &matrix)
{
    for(int i=0; i<matrix.size(); i++){
        printf("{");
        for(int j=0; j< matrix[i].size(); j++) {
            printf("%3d ", matrix[i][j]) ;
        }
        printf("}\n");
    }
    cout << endl;
}


int main()
{
    //int a[] = {-1, 0, 1, 2, -1, 1, -4};
    int a[] = {-1, 1, 1, 1, -1, -1, 0,0,0};
    vector<int> n(a, a+sizeof(a)/sizeof(int));
    vector< vector<int> > result = threeSum(n);
    printMatrix(result);    
    return 0;
}