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Maximum occurred integer in n ranges | Set-2
  • Last Updated : 16 Jun, 2020

Given N ranges of L-R. The task is to print the number which occurs the maximum number of times in the given ranges.

Note: 1 <= L <= R <= 106


Examples:

Input: range[] = { {1, 6}, {2, 3}, {2, 5}, {3, 8} }
Output: 3
1 occurs in 1 range {1, 6}
2 occurs 3 in 3 range {1, 6}, {2, 3}, {2, 5}
3 occurs 4 in 4 range {1, 6}, {2, 3}, {2, 5}, {3, 8}
4 occurs 3 in 3 range {1, 6}, {2, 5}, {3, 8}
5 occurs 3 in 3 range {1, 6}, {2, 5}, {3, 8}
6 occurs 2 in 2 range {1, 6}, {3, 8}
7 occurs 1 in 1 range {3, 8}
8 occurs 1 in 1 range {3, 8}

Input: range[] = { {1, 4}, {1, 9}, {1, 2}};
Output: 1



Approach: The approach is similar to Maximum occurred integer in n ranges. The only thing that is different is to find the lower and upper bound of ranges. So that there is no need to traverse from 1 to MAX.

Below is the step by step algorithm to solve this problem:

  1. Initialize a freq array with 0, let the size of the array be 10^6 as this is the maximum possible.
  2. Increase the freq[l] by 1, for every starting index of the given range.
  3. Decrease the freq[r+1] by 1 for every ending index of the given range.
  4. Iterate from the minimum L to the maximum R and add the frequencies by freq[i] += freq[i-1].
  5. The index with the maximum value of freq[i] will be the answer.
  6. Store the index and return it.

Below is the implementation of above approach:

C++




// C++ program to check the most occurring
// element in given range
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the maximum element.
int maxOccurring(int range[][2], int n)
{
  
    // freq array to store the frequency
    int freq[(int)(1e6 + 2)] = { 0 };
  
    int first = 0, last = 0;
  
    // iterate and mark the hash array
    for (int i = 0; i < n; i++) {
        int l = range[i][0];
        int r = range[i][1];
  
        // increase the hash array by 1 at L
        freq[l] += 1;
  
        // Decrease the hash array by 1 at R
        freq[r + 1] -= 1;
  
        first = min(first, l);
        last = max(last, r);
    }
  
    // stores the maximum frequency
    int maximum = 0;
    int element = 0;
  
    // check for the most occurring element
    for (int i = first; i <= last; i++) {
  
        // increase the frequency
        freq[i] = freq[i - 1] + freq[i];
  
        // check if is more than the previous one
        if (freq[i] > maximum) {
            maximum = freq[i];
            element = i;
        }
    }
  
    return element;
}
  
// Driver code
int main()
{
    int range[][2] = { { 1, 6 }, { 2, 3 }, { 2, 5 }, { 3, 8 } };
    int n = 4;
  
    // function call
    cout << maxOccurring(range, n);
  
    return 0;
}

Java




// Java program to check the most occurring
// element in given range
class GFG 
  
// Function that returns the maximum element.
static int maxOccurring(int range[][], int n)
{
  
    // freq array to store the frequency
    int []freq = new int[(int)(1e6 + 2)];
  
    int first = 0, last = 0;
  
    // iterate and mark the hash array
    for (int i = 0; i < n; i++)
    {
        int l = range[i][0];
        int r = range[i][1];
  
        // increase the hash array by 1 at L
        freq[l] += 1;
  
        // Decrease the hash array by 1 at R
        freq[r + 1] -= 1;
  
        first = Math.min(first, l);
        last = Math.max(last, r);
    }
  
    // stores the maximum frequency
    int maximum = 0;
    int element = 0;
  
    // check for the most occurring element
    for (int i = first+1; i <= last; i++) 
    {
  
        // increase the frequency
        freq[i] = freq[i - 1] + freq[i];
  
        // check if is more than the previous one
        if (freq[i] > maximum) 
        {
            maximum = freq[i];
            element = i;
        }
    }
    return element;
}
  
// Driver code
public static void main(String[] args) 
{
    int range[][] = { { 1, 6 }, { 2, 3 }, 
                      { 2, 5 }, { 3, 8 } };
    int n = 4;
  
    // function call
    System.out.println(maxOccurring(range, n));
}
}
  
// This code is contributed by PrinciRaj1992 

Python3




# Python3 program to check the most 
# occurring element in given range
  
# Function that returns the 
# maximum element.
def maxOccurring(range1, n):
      
    # freq array to store the frequency
    freq = [0] * 1000002;
      
    first = 0;
    last = 0;
      
    # iterate and mark the hash array
    for i in range(n):
        l = range1[i][0];
        r = range1[i][1];
          
        # increase the hash array by 1 at L
        freq[l] += 1;
          
        # Decrease the hash array by 1 at R
        freq[r + 1] -= 1;
        first = min(first, l);
        last = max(last, r);
      
    # stores the maximum frequency
    maximum = 0;
    element = 0;
      
    # check for the most occurring element
    for i in range(first, last + 1):
          
        # increase the frequency
        freq[i] = freq[i - 1] + freq[i];
          
        # check if is more than the 
        # previous one
        if(freq[i] > maximum):
            maximum = freq[i];
            element = i;
    return element;
  
# Driver code
range1= [[ 1, 6 ], [ 2, 3 ], 
         [ 2, 5 ], [ 3, 8 ]];
n = 4;
      
# function call
print(maxOccurring(range1, n));
  
# This code is contributed by mits

C#




// C# program to check the most occurring
// element in given range
using System;
      
class GFG 
  
// Function that returns the maximum element.
static int maxOccurring(int [,]range, int n)
{
  
    // freq array to store the frequency
    int []freq = new int[(int)(1e6 + 2)];
  
    int first = 0, last = 0;
  
    // iterate and mark the hash array
    for (int i = 0; i < n; i++)
    {
        int l = range[i, 0];
        int r = range[i, 1];
  
        // increase the hash array by 1 at L
        freq[l] += 1;
  
        // Decrease the hash array by 1 at R
        freq[r + 1] -= 1;
  
        first = Math.Min(first, l);
        last = Math.Max(last, r);
    }
  
    // stores the maximum frequency
    int maximum = 0;
    int element = 0;
  
    // check for the most occurring element
    for (int i = first + 1; i <= last; i++) 
    {
  
        // increase the frequency
        freq[i] = freq[i - 1] + freq[i];
  
        // check if is more than the previous one
        if (freq[i] > maximum) 
        {
            maximum = freq[i];
            element = i;
        }
    }
    return element;
}
  
// Driver code
public static void Main(String[] args) 
{
    int [,]range = {{ 1, 6 }, { 2, 3 }, 
                    { 2, 5 }, { 3, 8 }};
    int n = 4;
  
    // function call
    Console.WriteLine(maxOccurring(range, n));
}
}
  
// This code is contributed by Princi Singh

PHP




<?php 
// PHP program to check the most occurring
// element in given range
  
// Function that returns the maximum element.
function maxOccurring(&$range, $n)
{
    // freq array to store the frequency
    $freq = array(0, 1000002, NULL); 
  
    $first = 0;
    $last = 0;
  
    // iterate and mark the hash array
    for ($i = 0; $i < $n; $i++)
    {
        $l = $range[$i][0];
        $r = $range[$i][1];
  
        // increase the hash array 
        // by 1 at L
        $freq[$l] += 1;
  
        // Decrease the hash array 
        // by 1 at R
        $freq[$r + 1] -= 1;
  
        $first = min($first, $l);
        $last = max($last, $r);
    }
  
    // stores the maximum frequency
    $maximum = 0;
    $element = 0;
  
    // check for the most occurring element
    for ($i = $first; $i <= $last; $i++)
    {
  
        // increase the frequency
        $freq[$i] = $freq[$i - 1] + $freq[$i];
  
        // check if is more than the 
        // previous one
        if ($freq[$i] > $maximum)
        {
            $maximum = $freq[$i];
            $element = $i;
        }
    }
  
    return $element;
}
  
// Driver code
$range = array(array( 1, 6 ),
               array( 2, 3 ),
               array( 2, 5 ), 
               array( 3, 8 ));
$n = 4;
  
// function call
echo maxOccurring($range, $n);
  
// This code is contributed by ita_c
?>
Output:
3

Note: If values the values of L and T are of order 108 then above method won’t work as there will be a memory error. We need a different but similar approach for these limits. You may think in terms of hashing.

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