# Maximum number of overlapping Intervals

Given different intervals, the task is to print the maximum number of overlap among these intervals at any time.

Examples:

Input: v = {{1, 2}, {2, 4}, {3, 6}}
Output: 2
The maximum overlapping is 2(between (1 2) and (2 4) or between (2 4) and (3 6))

Input: v = {{1, 8}, {2, 5}, {5, 6}, {3, 7}}
Output: 4
The maximum overlapping is 4 (between (1, 8), (2, 5), (5, 6) and (3, 7))

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

Approach:

• The idea is to store coordinates in a new vector of pair mapped with characters ‘x’ and ‘y’, to identify coordinates.
• Sort the vector.
• Traverse the vector, if an x coordinate is encountered it means a new range is added, so update count and if y coordinate is encountered that means a range is subtracted.
• Update the value of count for every new coordinate and take maximum.

## CPP

 `// C++ program that print maximum ` `// number of overlap ` `// among given ranges ` `#include ` `using` `namespace` `std; ` ` `  `// Function that print maximum ` `// overlap among ranges ` `void` `overlap(vector > v) ` `{ ` `    ``// variable to store the maximum ` `    ``// count ` `    ``int` `ans = 0; ` `    ``int` `count = 0; ` `    ``vector > data; ` ` `  `    ``// storing the x and y ` `    ``// coordinates in data vector ` `    ``for` `(``int` `i = 0; i < v.size(); i++) { ` ` `  `        ``// pushing the x coordinate ` `        ``data.push_back({ v[i].first, ``'x'` `}); ` ` `  `        ``// pushing the y coordinate ` `        ``data.push_back({ v[i].second, ``'y'` `}); ` `    ``} ` ` `  `    ``// sorting of ranges ` `    ``sort(data.begin(), data.end()); ` ` `  `    ``// Traverse the data vector to ` `    ``// count number of overlaps ` `    ``for` `(``int` `i = 0; i < data.size(); i++) { ` ` `  `        ``// if x occur it means a new range ` `        ``// is added so we increase count ` `        ``if` `(data[i].second == ``'x'``) ` `            ``count++; ` ` `  `        ``// if y occur it means a range ` `        ``// is ended so we decrease count ` `        ``if` `(data[i].second == ``'y'``) ` `            ``count--; ` ` `  `        ``// updating the value of ans ` `        ``// after every traversal ` `        ``ans = max(ans, count); ` `    ``} ` ` `  `    ``// printing the maximum value ` `    ``cout << ans << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector > v ` `        ``= { { 1, 2 }, { 2, 4 }, { 3, 6 } }; ` `    ``overlap(v); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 program that prmaximum ` `# number of overlap ` `# among given ranges ` ` `  `# Function that prmaximum ` `# overlap among ranges ` `def` `overlap(v): ` ` `  `    ``# variable to store the maximum ` `    ``# count ` `    ``ans ``=` `0` `    ``count ``=` `0` `    ``data ``=` `[] ` ` `  `    ``# storing the x and y ` `    ``# coordinates in data vector ` `    ``for` `i ``in` `range``(``len``(v)): ` ` `  `        ``# pushing the x coordinate ` `        ``data.append([v[i][``0``], ``'x'``]) ` ` `  `        ``# pushing the y coordinate ` `        ``data.append([v[i][``1``], ``'y'``]) ` ` `  `    ``# sorting of ranges ` `    ``data ``=` `sorted``(data) ` ` `  `    ``# Traverse the data vector to ` `    ``# count number of overlaps ` `    ``for` `i ``in` `range``(``len``(data)): ` ` `  `        ``# if x occur it means a new range ` `        ``# is added so we increase count ` `        ``if` `(data[i][``1``] ``=``=` `'x'``): ` `            ``count ``+``=` `1` ` `  `        ``# if y occur it means a range ` `        ``# is ended so we decrease count ` `        ``if` `(data[i][``1``] ``=``=` `'y'``): ` `            ``count ``-``=` `1` ` `  `        ``# updating the value of ans ` `        ``# after every traversal ` `        ``ans ``=` `max``(ans, count) ` ` `  `    ``# printing the maximum value ` `    ``print``(ans) ` ` `  `# Driver code ` `v ``=` `[ [ ``1``, ``2` `], [ ``2``, ``4` `], [ ``3``, ``6` `] ] ` `overlap(v) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```2
```

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