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Count distinct points visited on the number line in C++
We are given a binary sequence of 0’s and 1’s. Also assume a person is sitting at a position or point stored in current_pos. Now starting from current_pos, if the binary sequence has 0 then he moves one step left ( current_pos - 1). If it’s 1 he moves one step right ( current_pos + 1). The goal is to find distinct positions or points he visited after the whole binary sequence is completed.
We will solve this using the number of times a point is visited. If frequency is non-zero, increase count of distinct points.
Input
Path[]= “001100” current_pos=3
Output
Distinct points visited on number line: 3
Explanation − starting from path[0] and position 3.
Path[0]: 0 → move left ...currpos=2 Path[1]: 0 → move left ...currpos=1 Path[2]: 1 → move right ...currpos=2 Path[3]: 1 → move left ...currpos=3 Path[4]: 0 → move left ...currpos=2 Path[5]: 0 → move left ...currpos=1
Total distinct positions are 3, which are 1,2,3
Input
Path[]= “010101” current_pos=5
Output
Distinct points visited on number line: 2
Explanation − starting from path[0] and position 5.
Path[0]: 0 → move left ...currpos=4 Path[1]: 1 → move right ...currpos=5 Path[2]: 0 → move left ...currpos=4 Path[3]: 1 → move right ...currpos=5 Path[4]: 0 → move left ...currpos=4 Path[5]: 1 → move right ...currpos=5
Total distinct positions are 2, which are 4 and 5
Approach used in the below program is as follows
String of 0’s and 1’s is stored in the path.
Current_pos stores starting point.
Function getDistinctPoints(int current_pos, string path) takes current position and path as input and returns the count of distinct positions/points.
Variable len stores the length of path.
Array frequency[21] is used to store the frequency for the number of times the point is visited. Index represents the point. Total 0-20.
Start traversing the path string.
If current value is 0, move left ( current_pos -1 ). Update frequency of this point frequency[current_pos]++.
Else current value is 1, move right ( current_pos +1 ). Update frequency of this point frequency[current_pos]++.
Now traverse the frequency array and for each non zero value. Increase count.
Count contains distinct visited points.
- Return count as desired result.
Example
#include <bits/stdc++.h>
using namespace std;
//distict points visited between 0 to 20
int getDistinctPoints(int current_pos, string path){
// Length of path
int len = path.length();
int count=0;
// Array to store number of times a point is visited
int frequency[21]={0};
// For all the directions in path
for (int i = 0; i < len; i++) {
//for left direction
if (path[i] == '0') {
current_pos--;
frequency[current_pos]++; //increase visit count
}
// If the direction is right
else {
current_pos++;
frequency[current_pos]++; //increase visit count
}
}
for(int i=0;i<21;i++)
if(frequency[i]!=0) // if visted then frequency[i] is non zero
count++;
return count;
}
int main(){
int current_pos = 3;
string path = "011101100";
cout << "Count of distinct points visited on the number line:<<(getDistinctPoints(current_pos, path));
return 0;
}Output
Count of distinct points visited on the number line :5
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