Interviewstreet-pairs-category search

Source of https://www.interviewstreet.com/challenges/dashboard/#problem/4e14b83d5fd12

Solution report:

It is the simplest algorithm used to sort the data first, and then traverse the data from the past to the next, calculate the value + k for each number, and then look for whether the value + k exists in the second part.

#include <iostream>#include <algorithm>using namespace std;int main(){    int N, K;    cin >> N >> K;    long long *key = new long long[N];    for (int i = 0; i < N; i++)    {        cin >> key[i];    }    sort (key, key+N);    int k = 0;    for (int i = 0; i < N; i++)    {        int begin = i+1;        int end = N-1;        int value = key[i] + K;        while(begin <= end)        {            int mid = (begin+end)/2;            if (key[mid] < value)                begin = mid + 1;            else if (key[mid] > value)                end = mid - 1;            else            {                k++;                break;            }        }    }    cout << k << endl;    }

Appendix:

Given n numbers, [n <= 10 ^ 5] We need to count the total pairs of numbers that have a difference of K. [k> 0 and K <1e9]

Input Format:
1st line contains N & K (integers ).
2nd line contains N numbers of the set. All the N numbers are assured to be distinct.
Output Format:
One integer saying the NO of pairs of numbers that have a diff K.

Sample input #00:
5 2
1 5 3 4 2

Sample output #00:
3

  Sample input #01:10 1363374326 364147530 61825163 1073065718 1281246024 1399469912 428047635 491595254 879792181 1069262793