Given an array of integers, the task is to find the minimum number of segments that the array elements can be divided into such that all the segments contain distinct elements.
Examples:
Input: n=6 ; Array: 1, 7, 4, 3, 3, 8
Output: 2
Explanation:
Optimal way to create segments here is {1, 7, 4, 3} {3, 8}
Clearly, the answer is the maximum frequency of any element within the array i.e. '2'.
as '3' is the element which appears the most in the array (twice).
Input : n=5 ; Array: 2, 2, 3, 3, 3, 5
Output : 3
Approach:
- The optimal approach is to put all the distinct elements in a single segment.
- And then, put all other elements that have several occurrences one by one in new segments such that no segment contains repetitions of elements.
- So, the answer is the maximum frequency of any element within the given array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that counts the // minimum segments required void CountSegments(int N, int a[]) { // all values are '0' initially int frequency[10001] = { 0 }; // count of segments int c = 0; // store frequency of every element for (int i = 0; i < N; i++) { frequency[a[i]]++; } // find maximum frequency for (int i = 0; i <= 10000; i++) c = max(c, frequency[i]); cout << c << "\n"; } // Driver code int main() { int N = 6; int a[] = { 1, 3, 4, 3, 2, 3 }; CountSegments(N, a); return 0; } |
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Java
// Java implementation of the approach import java.io.*; class GFG { // Function that counts the // minimum segments required static void CountSegments(int N, int a[]) { // all values are '0' initially int frequency[] = new int[10001]; // count of segments int c = 0; // store frequency of every element for (int i = 0; i < N; i++) { frequency[a[i]]++; } // find maximum frequency for (int i = 0; i <= 10000; i++) c = Math.max(c, frequency[i]); System.out.println( c); } // Driver code public static void main (String[] args) { int N = 6; int []a = { 1, 3, 4, 3, 2, 3 }; CountSegments(N, a); } } // This Code is contributed by inder_verma.. |
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Python 3
# Python3 implementation of the approach # Function that counts the # minimum segments required def CountSegments(N, a): # all values are '0' initially frequency = [0] * 10001 # count of segments c = 0 # store frequency of every element for i in range(N) : frequency[a[i]] += 1 # find maximum frequency for i in range(10001): c = max(c, frequency[i]) print(c) # Driver code if __name__ == "__main__": N = 6 a = [ 1, 3, 4, 3, 2, 3 ] CountSegments(N, a) # This code is contributed # by ChitraNayal |
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C#
// C# implementation of the approach using System; class GFG { // Function that counts the // minimum segments required static void CountSegments(int N, int []a) { // all values are '0' initially int []frequency = new int[10001]; // count of segments int c = 0; // store frequency of every element for (int i = 0; i < N; i++) { frequency[a[i]]++; } // find maximum frequency for (int i = 0; i <= 10000; i++) c = Math.Max(c, frequency[i]); Console.WriteLine( c); } // Driver code public static void Main () { int N = 6; int []a = { 1, 3, 4, 3, 2, 3 }; CountSegments(N, a); } } // This code is contributed // by inder_verma |
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Output:
3
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