# Maximum occurred integer in n ranges

Given n ranges of the form L and R, the task is to find the maximum occurred integer in all the ranges. If more than one such integer exists, print the smallest one.
0 <= Li, Ri < 1000000.

Examples :

```Input : L1 = 1 R1 = 15
L2 = 4 R2 = 8
L3 = 3 R3 = 5
L3 = 1 R3 = 4
Output : 4

Input : L1 = 1 R1 = 15
L2 = 5 R2 = 8
L3 = 9 R3 = 12
L4 = 13 R4 = 20
L5 = 21 R5 = 30
Output : 5
Numbers having maximum occurrence i.e 2  are 5, 6,
7, 8, 9, 10, 11, 12, 13, 14, 15. The smallest number
among all are 5.
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A simple solution is to use hash table to store counts of all numbers. We traverse every interval from Li to Ri and increment count of all numbers present in every interval. Finally we traverse the hash table to find the number with maximum count. Time complexity of this solution is O(n * MAX_INTERVAL) where MAX_INTERVAL is maximum number of elements in an interval.

An efficient solution requires linear time. We create an array arr[] of size 1000000 (limit given on maximum value of an interval). The idea is to add +1 to each Li index and -1 to corresponding Ri index in arr[]. After this, find the prefix sum of the array. Adding +1 at Li shows the starting point of ith Range and adding -1 shows the ending point of ith range. Finally we return the array index that has maximum prefix sum

Algorithm to solve the problem:

1. Initialize an array arr[] to 0.
2. For each range i, add +1 at Li index and -1 at Ri of the array.
3. Find the prefix sum of the array and find the smallest index having maximum prefix sum.

Below is the implementation of this approach:

## C++

 `// C++ program to find maximum occurred element in ` `// given N ranges. ` `#include ` `#define MAX 1000000 ` `using` `namespace` `std; ` ` `  `// Return the maximum occurred element in all ranges. ` `int` `maximumOccurredElement(``int` `L[], ``int` `R[], ``int` `n) ` `{ ` `    ``// Initalising all element of array to 0. ` `    ``int` `arr[MAX]; ` `    ``memset``(arr, 0, ``sizeof` `arr); ` ` `  `    ``// Adding +1 at Li index and substracting 1 ` `    ``// at Ri index. ` `    ``int` `maxi=-1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``arr[L[i]] += 1; ` `        ``arr[R[i] + 1] -= 1; ` `        ``if``(R[i]>maxi){ ` `            ``maxi=R[i]; ` `        ``} ` `    ``} ` ` `  `    ``// Finding prefix sum and index having maximum ` `    ``// prefix sum. ` `    ``int` `msum = arr,ind; ` `    ``for` `(``int` `i = 1; i < maxi+1; i++) { ` `        ``arr[i] += arr[i - 1]; ` `        ``if` `(msum < arr[i]) { ` `            ``msum = arr[i]; ` `            ``ind = i; ` `        ``} ` `    ``} ` ` `  `    ``return` `ind; ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `L[] = { 1, 4, 9, 13, 21 }; ` `    ``int` `R[] = { 15, 8, 12, 20, 30 }; ` `    ``int` `n = ``sizeof``(L) / ``sizeof``(L); ` ` `  `    ``cout << maximumOccurredElement(L, R, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find maximum occurred  ` `// element in given N ranges. ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `MAX = ``1000000``; ` ` `  `    ``// Return the maximum occurred element in all ranges. ` `    ``static` `int` `maximumOccurredElement(``int``[] L, ``int``[] R, ``int` `n) ` `    ``{ ` `        ``// Initalising all element of array to 0. ` `        ``int``[] arr = ``new` `int``[MAX]; ` ` `  `        ``// Adding +1 at Li index and  ` `        ``// substracting 1 at Ri index. ` `        ``int` `maxi=-``1``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``arr[L[i]] += ``1``; ` `            ``arr[R[i] + ``1``] -= ``1``; ` `            ``if``(R[i]>maxi){ ` `            ``maxi=R[i]; ` `           ``} ` `        ``} ` ` `  `        ``// Finding prefix sum and index ` `        ``// having maximum prefix sum. ` `        ``int` `msum = arr[``0``]; ` `        ``int` `ind = ``0``; ` `        ``for` `(``int` `i = ``1``; i < maxi+``1``; i++) { ` `            ``arr[i] += arr[i - ``1``]; ` `            ``if` `(msum < arr[i]) { ` `                ``msum = arr[i]; ` `                ``ind = i; ` `            ``} ` `        ``} ` ` `  `        ``return` `ind; ` `    ``} ` ` `  `    ``// Driver program ` `    ``static` `public` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] L = { ``1``, ``4``, ``9``, ``13``, ``21` `}; ` `        ``int``[] R = { ``15``, ``8``, ``12``, ``20``, ``30` `}; ` `        ``int` `n = L.length; ` `        ``System.out.println(maximumOccurredElement(L, R, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Python 3 program to find maximum occurred  ` `# element in given N ranges. ` ` `  `MAX` `=` `1000000` ` `  `# Return the maximum occurred element  ` `# in all ranges. ` `def` `maximumOccurredElement(L, R, n): ` `     `  `    ``# Initalising all element of array to 0. ` `    ``arr ``=` `[``0` `for` `i ``in` `range``(``MAX``)] ` ` `  `    ``# Adding +1 at Li index and substracting 1 ` `    ``# at Ri index. ` `    ``for` `i ``in` `range``(``0``, n, ``1``): ` `        ``arr[L[i]] ``+``=` `1` `        ``arr[R[i] ``+` `1``] ``-``=` `1` ` `  `    ``# Finding prefix sum and index ` `    ``# having maximum prefix sum. ` `    ``msum ``=` `arr[``0``] ` `    ``for` `i ``in` `range``(``1``, ``MAX``, ``1``): ` `        ``arr[i] ``+``=` `arr[i ``-` `1``] ` `        ``if` `(msum < arr[i]): ` `            ``msum ``=` `arr[i] ` `            ``ind ``=` `i ` `    ``return` `ind ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``L ``=` `[``1``, ``4``, ``9``, ``13``, ``21``] ` `    ``R ``=` `[``15``, ``8``, ``12``, ``20``, ``30``] ` `    ``n ``=` `len``(L) ` ` `  `    ``print``(maximumOccurredElement(L, R, n)) ` `     `  `# This code is contributed by ` `# Sanjit_Prasad `

## C#

 `// C# program to find maximum  ` `// occurred element in given N ranges. ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `MAX = 1000000; ` ` `  `    ``// Return the maximum occurred element in all ranges. ` `    ``static` `int` `maximumOccurredElement(``int``[] L, ``int``[] R, ``int` `n) ` `    ``{ ` `        ``// Initalising all element of array to 0. ` `        ``int``[] arr = ``new` `int``[MAX]; ` ` `  `        ``// Adding +1 at Li index and  ` `        ``// substracting 1 at Ri index. ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``arr[L[i]] += 1; ` `            ``arr[R[i] + 1] -= 1; ` `        ``} ` ` `  `        ``// Finding prefix sum and index  ` `        ``// having maximum prefix sum. ` `        ``int` `msum = arr; ` `        ``int` `ind = 0; ` `        ``for` `(``int` `i = 1; i < MAX; i++) { ` `            ``arr[i] += arr[i - 1]; ` `            ``if` `(msum < arr[i]) { ` `                ``msum = arr[i]; ` `                ``ind = i; ` `            ``} ` `        ``} ` ` `  `        ``return` `ind; ` `    ``} ` ` `  `    ``// Driver program ` `    ``static` `public` `void` `Main() ` `    ``{ ` `        ``int``[] L = { 1, 4, 9, 13, 21 }; ` `        ``int``[] R = { 15, 8, 12, 20, 30 }; ` `        ``int` `n = L.Length; ` `        ``Console.WriteLine(maximumOccurredElement(L, R, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```4
```

Time Complexity: O(n + MAX)

Exercise: Try for 0 <= Li, Ri <= 1000000000. (Hint: Use stl map).

Related Article : Maximum value in an array after m range increment operations

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