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:
- Initialize a freq array with 0, let the size of the array be 10^6 as this is the maximum possible.
- Increase the freq[l] by 1, for every starting index of the given range.
- Decrease the freq[r+1] by 1 for every ending index of the given range.
- Iterate from the minimum L to the maximum R and add the frequencies by freq[i] += freq[i-1].
- The index with the maximum value of freq[i] will be the answer.
- 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 ?> |
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. Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.