# Count the number of intervals in which a given value lies

• Difficulty Level : Medium
• Last Updated : 08 Jul, 2022

Given a 2D-array of integer intervals and a value . For every interval [li, ri], the task is to check if V lies in between li and ri, both inclusive. The task is to print the count of intervals that satisfies this.
Note: 1<=li<= ri
Examples:

Input: arr[][] = {{1, 10}, {5, 10}, {15, 25}, {7, 12}, {20, 25}}, V = 7
Output: 3
For V = 7, it lies in the intervals [1, 10], [5, 10] and [7, 12]. So, the count of intervals V lies in is 3.
Input: arr[][] = {{3, 7}, {2, 2}, {2, 8}, {7, 11}}, V = 2
Output: 2
For V = 2, it lies in the intervals [2, 2] and [2, 8]. So, the count of intervals V lies in is 2.

Method-1: Traverse the entire array checking for every interval [li, ri], if V lies in the interval or not. If it does, then increment the count.
Below is the implementation of the above approach:

## C++

 `// CPP program to count the``// number of intervals in which``// a given value lies` `#include``using` `namespace` `std;` `    ``// Function to count the``    ``// number of intervals in which``    ``// a given value lies``    ``int` `countIntervals(``int` `arr[], ``int` `V, ``int` `N)``    ``{``        ``// Variable to store the count of intervals``        ``int` `count = 0;` `        ``// Variables to store start and end of an interval``        ``int` `li, ri;` `        ``for` `(``int` `i = 0; i < N; i++)``       ``{``            ``li = arr[i];``            ``ri = arr[i];` `            ``// Implies V lies in the interval``            ``// so increase count``            ``if` `(V >= li && V <= ri)``                ``count++;``        ``}``        ``return` `count;``    ``}` `    ``// Driver code``    ``int` `main()``    ``{``        ``int` `arr[] = { { 1, 10 }, { 5, 10 },{ 15, 25 }, { 7, 12 }, { 20, 25 } };` `        ``int` `V = 7;` `        ``// length of the array``        ``int` `N = ``sizeof``(arr)/``sizeof``(arr);``        ``cout<<(countIntervals(arr, V, N))<

## Java

 `// Java program to count the``// number of intervals in which``// a given value lies` `import` `java.io.*;``import` `java.util.*;``import` `java.lang.*;` `class` `GFG {` `    ``// Function to count the``    ``// number of intervals in which``    ``// a given value lies``    ``static` `int` `countIntervals(``int``[][] arr, ``int` `V, ``int` `N)``    ``{``        ``// Variable to store the count of intervals``        ``int` `count = ``0``;` `        ``// Variables to store start and end of an interval``        ``int` `li, ri;` `        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``li = arr[i][``0``];``            ``ri = arr[i][``1``];` `            ``// Implies V lies in the interval``            ``// so increase count``            ``if` `(V >= li && V <= ri)``                ``count++;``        ``}``        ``return` `count;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int``[][] arr = { { ``1``, ``10` `}, { ``5``, ``10` `},``         ``{ ``15``, ``25` `}, { ``7``, ``12` `}, { ``20``, ``25` `} };` `        ``int` `V = ``7``;` `        ``// length of the array``        ``int` `N = arr.length;` `        ``System.out.println(countIntervals(arr, V, N));``    ``}``}`

## Python3

 `# Python 3 program to count the``# number of intervals in which``# a given value lies` `# Function to count the``# number of intervals in which``# a given value lies``def` `countIntervals(arr, V, N) :``    ` `    ``# Variable to store the``    ``# count of intervals``    ``count ``=` `0` `    ``# Variables to store start and``    ``# end of an interval``    ``for` `i ``in` `range``(N) :``            ` `        ``# Variables to store start and``        ``# end of an interval``        ``li ``=` `arr[i][``0``]``        ``ri ``=` `arr[i][``1``]` `        ``# Implies V lies in the interval``        ``# so increase count``        ``if` `(V >``=` `li ``and` `V <``=` `ri) :``            ``count ``+``=` `1``        ` `    ``return` `count;``    ` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[ [ ``1``, ``10` `], [ ``5``, ``10` `],``            ``[ ``15``, ``25` `], [ ``7``, ``12` `],``            ``[ ``20``, ``25` `] ]` `    ``V ``=` `7` `    ``# length of the array``    ``N ``=` `len``(arr)` `    ``print``((countIntervals(arr, V, N)))` `# This code is contributed by Ryuga`

## C#

 `// C# program to count the``// number of intervals in which``// a given value lies``using` `System;` `class` `GFG``{` `// Function to count the``// number of intervals in which``// a given value lies``static` `int` `countIntervals(``int``[,] arr, ``int` `V, ``int` `N)``{``    ``// Variable to store the count of intervals``    ``int` `count = 0;` `    ``// Variables to store start and end of an interval``    ``int` `li, ri;` `    ``for` `(``int` `i = 0; i < N ; i++)``    ``{``        ``li = arr[i, 0];``        ``ri = arr[i, 1];` `        ``// Implies V lies in the interval``        ``// so increase count``        ``if` `(V >= li && V <= ri)``            ``count++;``    ``}``    ``return` `count;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int``[,] arr = ``new` `int``[,]{ { 1, 10 }, { 5, 10 },``    ``{ 15, 25 }, { 7, 12 }, { 20, 25 } };` `    ``int` `V = 7;` `    ``// length of the array``    ``int` `N = arr.GetLength(0);` `    ``Console.WriteLine(countIntervals(arr, V, N));``}``}` `// This code is contributed by Akanksha Rai`

## PHP

 `= ``\$li` `&& ``\$V` `<= ``\$ri``)``            ``\$count``++;``    ``}``    ``return` `\$count``;``}` `// Driver code``\$arr` `= ``array``(``array``(1, 10),``             ``array``(5, 10),``             ``array``(15, 25),``             ``array``(7, 12),``             ``array``(20, 25));` `\$V` `= 7;` `// length of the array``\$N` `= sizeof(``\$arr``);` `echo` `countIntervals(``\$arr``, ``\$V``, ``\$N``);` `// This code is contributed``// by Akanksha Rai``?>`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N)

Auxiliary Space: O(1)

Method-2(Efficient for queries): Use a frequency array that keeps track of how many of the given intervals an element lies in.

1. For every interval [L, R], put freq[L] as freq[L] + 1 and freq[R+1] as freq[R + 1] – 1 indicating start and end of the interval.
2. Keep track of the overall minimum and maximum of the intervals.
3. Starting from minimum to maximum construct the frequency array from its previous element i.e.,
freq[i] = freq[i] + freq[i – 1].
4. Required count of intervals is then given by freq[V].

Below is the implementation of the above approach:

## C++

 `// C++ program to count the``// number of intervals in which``// a given value lies``#include``using` `namespace` `std;` `const` `int` `MAX_VAL = 200000;` `// Function to count the``// number of intervals in which``// a given value lies``int` `countIntervals(``int` `arr[], ``int` `V, ``int` `N)``{``    ``// Variables to store overall minimum and``    ``// maximum of the intervals``    ``int` `min = INT_MAX;``    ``int` `max = INT_MIN;` `    ``// Variables to store start and``    ``// end of an interval``    ``int` `li, ri;` `    ``// Frequency array to keep track of``    ``// how many of the given intervals``    ``// an element lies in``    ``int` `freq[MAX_VAL];` `    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``li = arr[i];``        ``freq[li] = freq[li] + 1;``        ``ri = arr[i];``        ``freq[ri + 1] = freq[ri + 1] - 1;` `        ``if` `(li < min)``            ``min = li;``        ``if` `(ri > max)``            ``max = ri;``    ``}` `    ``// Constructing the frequency array``    ``for` `(``int` `i = min; i <= max; i++)``        ``freq[i] = freq[i] + freq[i - 1];` `    ``return` `freq[V];``}` `// Driver code``int` `main()``{``    ``int` `arr = { { 1, 10 }, { 5, 10 },``                      ``{ 15, 25 }, { 7, 12 },``                      ``{ 20, 25 } };` `    ``int` `V = 7;` `    ``// length of the array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << (countIntervals(arr, V, N));``}` `// This code is contributed by``// Surendra_Gangwar`

## Java

 `// Java program to count the``// number of intervals in which``// a given value lies` `import` `java.io.*;``import` `java.util.*;``import` `java.lang.*;` `class` `GFG {` `    ``static` `final` `int` `MAX_VAL = ``200000``;` `    ``// Function to count the``    ``// number of intervals in which``    ``// a given value lies``    ``static` `int` `countIntervals(``int``[][] arr, ``int` `V, ``int` `N)``    ``{``        ``// Variables to store overall minimum and``        ``// maximum of the intervals``        ``int` `min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;` `        ``// Variables to store start and end of an interval``        ``int` `li, ri;` `        ``// Frequency array to keep track of``        ``// how many of the given intervals an element lies in``        ``int``[] freq = ``new` `int``[MAX_VAL];` `        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``li = arr[i][``0``];``            ``freq[li] = freq[li] + ``1``;``            ``ri = arr[i][``1``];``            ``freq[ri + ``1``] = freq[ri + ``1``] - ``1``;` `            ``if` `(li < min)``                ``min = li;``            ``if` `(ri > max)``                ``max = ri;``        ``}` `        ``// Constructing the frequency array``        ``for` `(``int` `i = min; i <= max; i++)``            ``freq[i] = freq[i] + freq[i - ``1``];` `        ``return` `freq[V];``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int``[][] arr = { { ``1``, ``10` `}, { ``5``, ``10` `},``         ``{ ``15``, ``25` `}, { ``7``, ``12` `}, { ``20``, ``25` `} };` `        ``int` `V = ``7``;` `        ``// length of the array``        ``int` `N = arr.length;` `        ``System.out.println(countIntervals(arr, V, N));``    ``}``}`

## Python3

 `# Python3 program to count the number of``# intervals in which a given value lies``MAX_VAL ``=` `200000` `# Function to count the number of``# intervals in which a given value lies``def` `countIntervals(arr, V, N):``    ` `    ``# Variables to store overall minimumimum``    ``# and maximumimum of the intervals``    ``minimum ``=` `float``(``"inf"``)``    ``maximum ``=` `0` `    ``# Frequency array to keep track of``    ``# how many of the given intervals``    ``# an element lies in``    ``freq ``=` `[``0``] ``*` `(MAX_VAL)` `    ``for` `i ``in` `range``(``0``, N):``        ``li ``=` `arr[i][``0``]``        ``freq[li] ``=` `freq[li] ``+` `1``        ``ri ``=` `arr[i][``1``]``        ``freq[ri ``+` `1``] ``=` `freq[ri ``+` `1``] ``-` `1` `        ``if` `li < minimum:``            ``minimum ``=` `li``        ` `        ``if` `ri > maximum:``            ``maximum ``=` `ri``        ` `    ``# Constructing the frequency array``    ``for` `i ``in` `range``(minimum, maximum ``+` `1``):``        ``freq[i] ``=` `freq[i] ``+` `freq[i ``-` `1``]` `    ``return` `freq[V]``    ` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``arr ``=` `[[``1``, ``10``], [``5``, ``10``], [``15``, ``25``],``           ``[``7``, ``12``], [``20``, ``25``]]` `    ``V ``=` `7` `    ``# length of the array``    ``N ``=` `len``(arr)` `    ``print``(countIntervals(arr, V, N))` `# This code is contributed``# by Rituraj Jain`

## C#

 `// C# program to count the``// number of intervals in which``// a given value lies``using` `System;` `class` `GFG {` `    ``static`  `int` `MAX_VAL = 200000;` `    ``// Function to count the``    ``// number of intervals in which``    ``// a given value lies``    ``static` `int` `countIntervals(``int``[,] arr, ``int` `V, ``int` `N)``    ``{``        ``// Variables to store overall minimum and``        ``// maximum of the intervals``        ``int` `min = ``int``.MaxValue, max = ``int``.MinValue;` `        ``// Variables to store start and end of an interval``        ``int` `li, ri;` `        ``// Frequency array to keep track of``        ``// how many of the given intervals an element lies in``        ``int``[] freq = ``new` `int``[MAX_VAL];` `        ``for` `(``int` `i = 0; i < N; i++) {``            ``li = arr[i,0];``            ``freq[li] = freq[li] + 1;``            ``ri = arr[i,1];``            ``freq[ri + 1] = freq[ri + 1] - 1;` `            ``if` `(li < min)``                ``min = li;``            ``if` `(ri > max)``                ``max = ri;``        ``}` `        ``// Constructing the frequency array``        ``for` `(``int` `i = min; i <= max; i++)``            ``freq[i] = freq[i] + freq[i - 1];` `        ``return` `freq[V];``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[,] arr = ``new` `int``[,]{ { 1, 10 }, { 5, 10 },``        ``{ 15, 25 }, { 7, 12 }, { 20, 25 } };` `        ``int` `V = 7;` `        ``// length of the array``        ``int` `N = arr.Length/arr.Rank;` `        ``Console.WriteLine(countIntervals(arr, V, N));``    ``}``}``// this code is contributed by mits`

## PHP

 ` ``\$max``)``            ``\$max` `= ``\$ri``;``    ``}` `    ``// Constructing the frequency array``    ``for` `(``\$i` `= ``\$min``; ``\$i` `<= ``\$max``; ``\$i``++)``        ``\$freq``[``\$i``] = ``\$freq``[``\$i``] + ``\$freq``[``\$i` `- 1];` `    ``return` `\$freq``[``\$V``];``}` `// Driver code``\$arr` `= ``array``(``array``( 1, 10 ),``             ``array``( 5, 10 ),``             ``array``( 15, 25 ),``             ``array``( 7, 12 ),``             ``array``( 20, 25 ));` `\$V` `= 7;` `// length of the array``\$N` `= ``count``(``\$arr``);` `echo` `(countIntervals(``\$arr``, ``\$V``, ``\$N``));` `// This code is contributed by mits``?>`

## Javascript

 ``

Output:

`3`

Time Complexity: O(n)

Auxiliary Space: O(MAX)

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