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Maximum number of segments that can contain the given points in C++
Given the task is to find the maximum of segments that can contain the given points.
Given an array a1[] with size n1 and two integers A and B are given. From the given array a1[], n1 line segments can be formed with starting and ending points as a1[i] – A and a1[i] + B respectively.
Another array a2[] is given with n2 number of points. These points have to be assigned to the line segments such that the number of line segments than have been assigned a point is maximum. Note that a single point can be assigned only once to a given line segment.
Let’s now understand what we have to do using an example:
Input
a1[] = {1, 4, 5}, a2[] = {2, 8}, A = 1, B = 2Output
1
Explanation − The line segments that can be formed using points a1[i] – A and a1[i] + B are (0, 6) and (3, 7).
The first point in the array a2[], that is, 2 can be assigned to the first line segment while the next point 8 cannot be assigned to any line segment. Therefore, only 1 line segment can be assigned a point and the output becomes 1.
Input
a1[] = {1, 2, 3, 4, 6, 7}, a2[] = {2, 5, 6, 8}, A = 0, B = 1Output
4
Approach used in the below program as follows
Initialize vectors a1 and a2 and integers A and B in the main function with certain values.
Create variable n1 and n2 and store in them, the size of vectors a1 and a2 respectively.
In Max() function first sort both the vectors a1 and a2.
Initialize j = 0 and ans = 0 for keeping track of vector a2 and the final answer respectively.
Loop from i = 0 till i < n1.
Inside the For loop initiate another while loop with condition j < n2.
Check if (a1[i] + B < a2[j]). If so then break out of the while loop.
Else check if (a2[j] >= a1[i] - A && a2[j] <= a1[i] + B). If so then increment ans and j and break out of the while loop.
If none of the above statement is true then just increment j.
- return ans
Example
#include <bits/stdc++.h>
using namespace std;
int Max(vector<int> a1, vector<int> a2, int n1, int n2, int A, int B){
//sorting a1 and a2
sort(a1.begin(), a1.end());
sort(a2.begin(), a2.end());
int j = 0;
int ans = 0;
for (int i = 0; i < n1; i++){
// searching for point
while (j < n2){
/* If ending point of segment is
smaller than the current point*/
if (a1[i] + B < a2[j])
break;
//
if (a2[j] >= a1[i] - A && a2[j] <= a1[i] + B){
ans++;
j++;
break;
}
else
j++;
}
}
return ans;
}
// main function
int main(){
int A = 0, B = 1;
vector<int> a1 = { 1, 2, 3, 4, 6, 7 };
int n1 = a1.size();
vector<int> a2 = { 2, 5, 6, 8 };
int n2 = a2.size();
cout << Max(a1, a2, n1, n2, A, B);
return 0;
}Output
4
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