Given an unsorted array of integers which may contain repeated elements, sort the elements in descending order of some of its occurrence. If there exists more than one element whose sum of occurrences are the same then, the one which is greater will come first.
Examples:
Input: arr[] = [2, 4, 1, 2, 4, 2, 10]
Output: arr[] = [10, 4, 4, 2, 2, 2, 1]Explanation:
Here, 2 appears 3 times, 4 appears 2 times, 1 appears 1 time and 10 appears 1 time.
Thus,
Sum of all occurrences of 2 in given array = 2 * 3 = 6,
Sum of all occurrences of 4 = 4 * 2 = 8,
Sum of all occurrences of 10 = 10 * 1 = 10,
Sum of all occurrences of 1 = 1 * 1 = 1.
Therefore sorting the array in descending order based on the above sum = [ 10, 4, 4, 2, 2, 2, 1 ]Input2: arr[] = [2, 3, 2, 3, 2, 1, 1, 9, 6, 9, 1, 2]
Output2: [9, 9, 2, 2, 2, 2, 6, 3, 3, 1, 1, 1]
Approach:
- Create a map that will map elements to it appearence ie if a occur one time then a will map to a but if a occur m times then a will be map to a*m.
- After doing the mapping sort the dictionary in descending order based on their values not keys. In case of tie, sort based on values.
- Finaly copy the sorted dictionary keys in final output array and there frequency is obtain based on their key-value pairs i.e. if a is mapped to a*m it means we need to include a, m times in output array.
Below is the implementation of the above approach:
Python
# Python3 program to sort elements of # arr[] in descending order of sum # of its occurrence def sort_desc(arr): # to store sum of all # occurrences of an elements d_sum = {} # to store count of # occurrence of elements d_count ={} # to store final result ans = [] # to store maximum sum of occurrence mx = 0 # Traverse and calculate sum # of occurrence and count of # occurrence of elements of arr[] for x in arr: if x not in d_sum: d_sum[x] = x d_count[x] = 1 else: d_sum[x] += x d_count[x] += 1 # sort d_sum in decreasing order of its value l = sorted(d_sum.items(), reverse = True, key = lambda x:(x[1], x[0])) # store the final result for x in l: ans += [x[0]] * d_count[x[0]] return ans # Driver Code arr = [3, 5, 2, 2, 3, 1, 3, 1] print(sort_desc(arr)) |
[3, 3, 3, 5, 2, 2, 1, 1]
Time Complexity: O(N*log N).
Space Complexity: O(N).
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