Beginner
1. Arithmetic Progression2. Geometric Progression3. Harmonic Progression4. No of triangles 5. Area of Polygon6. Trailing Zeros7. Min Platforms 8. Trapping Rain 9. Jump numbers10. Segregation of 0’s and 1’s11. Two SUM12. Triplet sum 13. Diagonal Sum14. Addition of two matrixes15. Subtraction of two matrixes16. Multiplication of two matrixes17. Transpose of given array18. Spiral Matrix19. Most Frequent Element20. Majority Element 21. Max difference 22. Odd occurring23. Second Largest and Second Smallest24. Next greater number25. Count Unique Elements26. Common elements 27. Removing Duplicates28. Balanced Array29. Rotating Array by k times30. Reverse Array31. Wave Array32. Merge arrays 33. Constraint Sort34. Find element in Sorted rotated array35. Sub Array Sum36. Largest subarray of 0’s and 1’s 37. K-Sub Array38. Max subarray sum39. Contiguous Square SumMost Frequent Element
LEVEL:Beginner
Description
Given list of elements. You have to find the most frequent element in given array. If there are more than two frequency elements then print the first occurrence in the given array.
Input Format
First line contains a number N. Next line contains N integers separated by spaces
Output Format
Print the resultant sum value.
5 3 5 4 4 1
4
8 5 4 3 6 5 4 3 4
4
10 3 2 5 4 6 7 6 7 6 5
6
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Approach
Approach 1: using brute force
Step-i) initialize two variables prev_count and curr_count to zero
Step-ii) initialize a variable with first element of the array assuming for now that it is the most frequent element
Step-iii) take the current element and check how many times it is repeated in the array
Step-iv) every time it is repeated increment the curr_count
Step-v) after checking entire array for that number check if curr_count is greater than prev_count
Step-vi) if it is greater assign curr_count to prev_count and answer variable to the current element
Step-vii)initialize curr_count to 0
Step-viii) repeat steps iii,iv,v,vi,vii till all the elements of the array are traversed
Step-ix) return the answer
Time Complexity: O(n^2)
Space Complexity:O(1)
Approach 2: using HashMap
Step-i) create a hashmap of both keys and values as integers
Step-ii)assign current element of the array to temp
Step-iii) if temp is in keys of hashmap increment its key value
Step-iv) if temp is not in keys of hashmap add current element as key and 1 as its value
Step-v)go to next element
Step-vi) repeat steps ii, iii, iv and v till you traverse the complete array
Step-vii) create a iterator for the hashmap
Step-viii) iterate over the hashmap and get the key of the highest value present in it
Step-ix) return the key
Time Complexity: O(n)
Space Complexity: O(n) as hashmap need to be maintained
Note :
Let us know if you can come up with a better approach, mail us at support@theinquisitive.in Your approach will be reviewed and posted with credits to you.
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