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Check one given element of an array present more than n/2 times or not.
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The elements of the array are taken as input. One value is taken from the user and that value is searched in the array whether it appears more than n/2 times in the array or not, where n is the number of elements of the array.
Algorithm
INPUT: Array elements
OUTPUT: the frequency of an element more than n/2 or not
PROCESS:
Step 1: [taking the input]
Read n [number of elements]
For i=0 to n-1 repeat
Read a[i]
[end of ‘for’ loop]
Read x [the element to search]
Step 2: [calculating the frequency of the element in array]
Set c<-0
For i=0 to n-1 repeat
If a[i]=x then
Set c<-c+1
[End of ‘if’]
[End of ‘for’ loop]
If c>n/2 then
Print “Yes, the element x is present more than n/2 times in the array”
Else
Print “No, the element x does not occur more than n/2 times in the array”
[End of ‘if’]
Step 3: Stop.
Code
Complexity:
The time complexity to take the number of elements is O(n) and the time complexity to find the frequency is also O(n).
Therefore, the time complexity of this program is O(n).
