# Find the missing elements from 1 to M in given N ranges

Given segments as ranges [L, R] where ranges are non-intersecting and non-overlapping. The task is to find all number between 1 to that doesn’t belong to any of the given ranges.

Examples:

```Input : N = 2, M = 6
Ranges:
[1, 2]
[4, 5]
Output : 3, 6
Explanation: Only 3 and 6 are missing from
the above ranges.

Input : N = 1, M = 5
Ranges:
[2, 4]
Output : 1, 5
```

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

Approach: Given that we have ranges, which are non-overlapping and non-intersecting. First of all, sort all segments based on starting value. After sorting, iterate from each segment and find the numbers which are missing.

Below is the implementation of the above approach:

## C++

 `// C++ program to find mssing elements ` `// from given Ranges ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find mssing elements ` `// from given Ranges ` `void` `findMissingNumber(vector > ranges, ``int` `m) ` `{ ` `    ``// First of all sort all the given ranges ` `    ``sort(ranges.begin(), ranges.end()); ` ` `  `    ``// store ans in a different vector ` `    ``vector<``int``> ans; ` ` `  `    ``// prev is use to store end of ` `    ``// last range ` `    ``int` `prev = 0; ` ` `  `    ``// j is used as a counter for ranges ` `    ``for` `(``int` `j = 0; j < ranges.size(); j++) { ` `        ``int` `start = ranges[j].first; ` `        ``int` `end = ranges[j].second; ` ` `  `        ``for` `(``int` `i = prev + 1; i < start; i++) ` `            ``ans.push_back(i); ` ` `  `        ``prev = end; ` `    ``} ` ` `  `    ``// for last segment ` `    ``for` `(``int` `i = prev + 1; i <= m; i++) ` `        ``ans.push_back(i); ` ` `  `    ``// finally print all answer ` `    ``for` `(``int` `i = 0; i < ans.size(); i++) { ` `        ``if` `(ans[i] <= m) ` `            ``cout << ans[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 2, M = 6; ` ` `  `    ``// Store ranges in vector of pair ` `    ``vector > ranges; ` `    ``ranges.push_back({ 1, 2 }); ` `    ``ranges.push_back({ 4, 5 }); ` ` `  `    ``findMissingNumber(ranges, M); ` ` `  `    ``return` `0; ` `} `

“”

## Python3

 `# Python3 program to find missing  ` `# elements from given Ranges  ` ` `  `# Function to find mssing elements  ` `# from given Ranges  ` `def` `findMissingNumber(ranges, m):  ` ` `  `    ``# First of all sort all the  ` `    ``# given ranges  ` `    ``ranges.sort()  ` ` `  `    ``# store ans in a different vector  ` `    ``ans ``=` `[]  ` ` `  `    ``# prev is use to store end ` `    ``# of last range ` `    ``prev ``=` `0` ` `  `    ``# j is used as a counter for ranges  ` `    ``for` `j ``in` `range``(``len``(ranges)):  ` `        ``start ``=` `ranges[j][``0``]  ` `        ``end ``=` `ranges[j][``1``]  ` ` `  `        ``for` `i ``in` `range``(prev ``+` `1``, start):  ` `            ``ans.append(i)  ` ` `  `        ``prev ``=` `end  ` ` `  `    ``# for last segment  ` `    ``for` `i ``in` `range``(prev ``+` `1``, m ``+` `1``):  ` `        ``ans.append(i)  ` ` `  `    ``# finally print all answer  ` `    ``for` `i ``in` `range``(``len``(ans)):  ` `        ``if` `ans[i] <``=` `m: ` `            ``print``(ans[i], end ``=` `" "``)  ` `     `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``N, M ``=` `2``, ``6` ` `  `    ``# Store ranges in vector of pair  ` `    ``ranges ``=` `[]  ` `    ``ranges.append([``1``, ``2``])  ` `    ``ranges.append([``4``, ``5``])  ` ` `  `    ``findMissingNumber(ranges, M)  ` ` `  `# This code is contributed  ` `# by Rituraj Jain `

## PHP

 ` `

Output:

```3 6
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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