Next Greater Frequency Element
Given an array, for each element find the value of nearest element to the right which is having frequency greater than as that of current element. If there does not exist an answer for a position, then make the value ‘1’.
Examples:
Input : a[] = [1, 1, 2, 3, 4, 2, 1] Output : [1, 1, 1, 2, 2, 1, 1] Explanation: Given array a[] = [1, 1, 2, 3, 4, 2, 1] Frequency of each element is: 3, 3, 2, 1, 1, 2, 3 Lets calls Next Greater Frequency element as NGF 1. For element a[0] = 1 which has a frequency = 3, As it has frequency of 3 and no other next element has frequency more than 3 so '1' 2. For element a[1] = 1 it will be 1 same logic like a[0] 3. For element a[2] = 2 which has frequency = 2, NGF element is 1 at position = 6 with frequency of 3 > 2 4. For element a[3] = 3 which has frequency = 1, NGF element is 2 at position = 5 with frequency of 2 > 1 5. For element a[4] = 4 which has frequency = 1, NGF element is 2 at position = 5 with frequency of 2 > 1 6. For element a[5] = 2 which has frequency = 2, NGF element is 1 at position = 6 with frequency of 3 > 2 7. For element a[6] = 1 there is no element to its right, hence 1 Input : a[] = [1, 1, 1, 2, 2, 2, 2, 11, 3, 3] Output : [2, 2, 2, 1, 1, 1, 1, 3, 1, 1]
Naive approach:
A simple hashing technique is to use values as index is be used to store frequency of each element. Create a list suppose to store frequency of each number in the array. (Single traversal is required). Now use two loops.
The outer loop picks all the elements one by one.
The inner loop looks for the first element whose frequency is greater than the frequency of current element.
If a greater frequency element is found then that element is printed, otherwise 1 is printed.
Time complexity : O(n*n)
Efficient approach:
We can use hashing and stack data structure to efficiently solve for many cases. A simple hashing technique is to use values as index and frequency of each element as value. We use stack data structure to store position of elements in the array.
1) Create a list to to use values as index to store frequency of each element.
2) Push the position of first element to stack.
3) Pick rest of the position of elements one by one and follow following steps in loop.
…….a) Mark the position of current element as ‘i’ .
……. b) If the frequency of the element which is pointed by the top of stack is greater than frequency of the current element, push the current position i to the stack
……. c) If the frequency of the element which is pointed by the top of stack is less than frequency of the current element and the stack is not empty then follow these steps:
…….i) continue popping the stack
…….ii) if the condition in step c fails then push the current position i to the stack
4) After the loop in step 3 is over, pop all the elements from stack and print 1 as next greater frequency element for them does not exist.
Time complexity is O(n).
Below is the Python 3 implementation of the above problem.

'''NFG function to find the next greater frequency
element for each element in the array'''
def
NFG(a, n):
if
(n <
=
0
):
print
(
"List empty"
)
return
[]
# stack data structure to store the position
# of array element
stack
=
[
0
]
*
n
# freq is a dictionary which maintains the
# frequency of each element
freq
=
{}
for
i
in
a:
freq[a[i]]
=
0
for
i
in
a:
freq[a[i]]
+
=
1
# res to store the value of next greater
# frequency element for each element
res
=
[
0
]
*
n
# initialize top of stack to 1
top
=

1
# push the first position of array in the stack
top
+
=
1
stack[top]
=
0
# now iterate for the rest of elements
for
i
in
range
(
1
, n):
''' If the frequency of the element which is
pointed by the top of stack is greater
than frequency of the current element
then push the current position i in stack'''
if
(freq[a[stack[top]]] > freq[a[i]]):
top
+
=
1
stack[top]
=
i
else
:
''' If the frequency of the element which
is pointed by the top of stack is less
than frequency of the current element, then
pop the stack and continuing popping until
the above condition is true while the stack
is not empty'''
while
(top>

1
and
freq[a[stack[top]]] < freq[a[i]]):
res[stack[top]]
=
a[i]
top

=
1
# now push the current element
top
+
=
1
stack[top]
=
i
'''After iterating over the loop, the remaining
position of elements in stack do not have the
next greater element, so print 1 for them'''
while
(top >

1
):
res[stack[top]]
=

1
top

=
1
# return the res list containing next
# greater frequency element
return
res
# Driver program to test the function
print
(NFG([
1
,
1
,
2
,
3
,
4
,
2
,
1
],
7
))
Output:
[1, 1, 1, 2, 2, 1, 1]
Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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