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Kth Smallest Number in Multiplication Table in C++
Suppose we know about one Multiplication Table. But could we find out the k-th smallest number quickly from the multiplication table? So if we have to height m and the length n of a m * n Multiplication Table, and one positive integer k, we have need to find the k-th smallest number in this table.
So if m = 3 and n = 3 and k is 6, then the output will be 4., this is because the multiplication table is like −
| 1 | 2 | 3 | |
| 1 | 1 | 2 | 3 |
| 2 | 2 | 4 | 6 |
| 3 | 3 | 6 | 9 |
6th smallest element is 4 as [1,2,2,3,3,4,6,6,9]
To solve this, we will follow these steps −
- Define a function ok(), this will take m, n, x,
- ret := 0
- for initialize i := 1, when i <= n, update (increase i by 1), do −
- temp := minimum of x / i and m
- ret := ret + temp
- return ret
- From the main method, do the following −
- ret := -1, low := 1, high := m * n
- while low <= high, do −
- mid := low + (high - low) / 2
- cnt := ok(m, n, mid)
- if cnt >= k, then −
- high := mid - 1
- ret := mid
- Otherwise
- low := mid + 1
- return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int ok(int m, int n, int x){
int ret = 0;
for(int i = 1; i <= n; i++){
int temp = min(x / i, m);
ret += temp;
}
return ret;
}
int findKthNumber(int m, int n, int k) {
int ret = -1;
int low = 1;
int high = m * n ;
while(low <= high){
int mid = low + (high - low)/ 2;
int cnt = ok(m, n, mid);
if(cnt >= k){
high = mid - 1;
ret = mid;
}else low = mid + 1;
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.findKthNumber(3,3,6));
}
Input
“2*”
Output
4
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