# Find number of segments covering each point in an given array

Given segments and some points, for each point find the number of segments covering that point.

A segment (l, r) covers a point x if and only if l < = x < = r.

Examples:

Input: Segments = {{0, 3}, {1, 3}, {3, 8}},
Points = {-1, 3, 8}.

Output : {0, 3, 1}

Explanation : • No segments passing through point -1
• All the segments passing through point 3
• Segment 3rd passing through point 8

Input: Segments = {{1, 3}, {2, 4}, {5, 7}},
Points = {0, 2, 5}.

Output: {0, 2, 1}

Explanation : • No segments passing through point 0
• 1st and 2nd segment passing through point 2
• Segment 3rd passing through point 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• We can do this by using logic similar to prefix sum.
• Let’s represent a segment with (l, r). Form a vector of pairs, for each segment push two pairs in vector with values (l, +1) ans (r + 1, -1).
• Sort the points in ascending order, but we also need it’s position so mapped it with it’s position.
• Sort the segment vector in descending order because we iterate on it from back.
• Make a variable count of segments, which is initially zero.
• Then, we will iterate on the point and pop the pair from the segment vector until it’s first value is less than equal to current point and add it’s second value to the count.
• Finally, Store the values of count in an array to his respective position and print the array.

Below is the implementation of the above approach.

## C++

 `// C++ program to find the number of  ` `// segments covering each points ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print an array ` `void` `PrintArray(``int` `n,``int` `arr[]) ` `{ ` `     ``for``(``int` `i = 0; i < n; i++)  ` `     ``{ ` `         ``cout< >segments, ` `                      ``vector<``int``>points, ``int` `s, ``int` `p) ` `{ ` `   ``vector< pair<``int``, ``int``> >pts, seg; ` `     `  `   ``// Pushing points and index in ` `   ``// vector as a pairs ` `   ``for``(``int` `i = 0; i < p; i++) ` `   ``{ ` `      ``pts.push_back({points[i], i});; ` `   ``} ` `     `  `   ``for``(``int` `i = 0; i < s; i++) ` `   ``{ ` `      ``// (l,+1) ` `      ``seg.push_back({segments[i].first, 1}); ` `      ``// (r+1,-1) ` `      ``seg.push_back({segments[i].second+1, -1}); ` `   ``} ` `     `  `   ``// Sort the vectors ` `   ``sort(seg.begin(), seg.end(),  ` `        ``greater>()); ` `   ``sort(pts.begin(),pts.end()); ` `     `  `   ``int` `count = 0; ` `   ``int` `ans[p]; ` `     `  `   ``for``(``int` `i = 0; i < p; i++) ` `   ``{ ` `        ``int` `x = pts[i].first; ` `        `  `        ``while``(!seg.empty() && ` `              ``seg.back().first <= x) ` `        ``{ ` `            ``count+= seg.back().second; ` `            ``seg.pop_back(); ` `        ``} ` `        ``ans[pts[i].second] = count; ` `   ``} ` `     `  `   ``// Print the answer ` `   ``PrintArray(p, ans); ` `   `  `} ` ` `  `//Driver code ` `int` `main() ` `{ ` `  ``// Initializing vector of pairs ` `  ``vector>seg; ` `     `  `  ``// Push segments ` `  ``seg.push_back({0, 3}); ` `  ``seg.push_back({1, 3}); ` `  ``seg.push_back({3, 8}); ` `     `  `  ``// Given points ` `  ``vector<``int``>point{-1, 3, 7}; ` `     `  `  ``int` `s = seg.size(); ` `  ``int` `p = point.size(); ` `     `  `  ``NumberOfSegments(seg, point, s, p); ` `     `  `  ``return` `0; ` `}  `

## Python3

 `# Python3 program to find the number  ` `# of segments covering each point ` ` `  `# Function to print an array ` `def` `PrintArray(n, arr): ` `     `  `    ``for` `i ``in` `range``(n): ` `        ``print``(arr[i], end ``=` `" "``) ` ` `  `# Funtion prints number of segments ` `# covering by each points ` `def` `NumberOfSegments(segments, points, s, p): ` `     `  `    ``pts ``=` `[] ` `    ``seg ``=` `[] ` `     `  `    ``# Pushing points and index in ` `    ``# vector as a pairs ` `    ``for` `i ``in` `range``(p): ` `        ``pts.append([points[i], i]) ` ` `  `    ``for` `i ``in` `range``(s): ` `         `  `        ``# (l, +1) ` `        ``seg.append([segments[i][``0``], ``1``]) ` `         `  `        ``# (r+1, -1) ` `        ``seg.append([segments[i][``1``] ``+` `1``, ``-``1``]) ` ` `  `    ``# Sort the vectors ` `    ``seg.sort(reverse ``=` `True``) ` `    ``pts.sort(reverse ``=` `False``) ` `     `  `    ``count ``=` `0` `    ``ans ``=` `[``0` `for` `i ``in` `range``(p)] ` ` `  `    ``for` `i ``in` `range``(p): ` `        ``x ``=` `pts[i][``0``] ` ` `  `        ``while``(``len``(seg) !``=` `0` `and`  `          ``seg[``len``(seg) ``-` `1``][``0``] <``=` `x): ` `                   `  `            ``count ``+``=` `seg[``len``(seg) ``-` `1``][``1``] ` `            ``seg.remove(seg[``len``(seg) ``-` `1``]) ` `             `  `        ``ans[pts[i][``1``]] ``=` `count ` `         `  `    ``# Print the answer ` `    ``PrintArray(p, ans) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``# Initializing vector of pairs ` `    ``seg ``=` `[] ` ` `  `    ``# Push segments ` `    ``seg.append([ ``0``, ``3` `]) ` `    ``seg.append([ ``1``, ``3` `]) ` `    ``seg.append([ ``3``, ``8` `]) ` `         `  `    ``# Given points ` `    ``point ``=` `[ ``-``1``, ``3``, ``7` `] ` `         `  `    ``s ``=` `len``(seg) ` `    ``p ``=` `len``(point) ` `         `  `    ``NumberOfSegments(seg, point, s, p) ` ` `  `# This code is contributed by Bhupendra_Singh `

Output:

```0 3 1
```

Time Complexity: O(s + p) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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