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; } |
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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 |
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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 |
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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 |
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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 ?> |
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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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