# Program to find most occurring number after k increments in python

PythonServer Side ProgrammingProgramming

#### Beyond Basic Programming - Intermediate Python

Most Popular

36 Lectures 3 hours

#### Practical Machine Learning using Python

Best Seller

91 Lectures 23.5 hours

#### Practical Data Science using Python

22 Lectures 6 hours

Suppose we have a list of numbers called nums and another value k. Let us consider an operation where we increase some element by one. We can perform at most k times, we have to find the value of the most frequently occurring number we can obtain. If there are more than one solution, select the smallest possible number.

So, if the input is like nums = [1, 0, 0, 0, 8, 8, 8, 8] k = 8, then the output will be 8, as we can increase 1, 7 times to get 8, and increase any 0 to 1, so, we get [8, 1, 0, 0, 8, 8, 8, 8]. So the result is 8.

To solve this, we will follow these steps:

• sort the list nums
• low := 0, high := 0
• dist := 0, best := 0
• ret := -1
• while high < size of nums, do
• if high > 0 and nums[high] is not same as nums[high - 1], then
• dist := dist +(high - low) *(nums[high] - nums[high - 1])
• high := high + 1
• while dist > k, do
• dist := dist - nums[high - 1] - nums[low]
• low := low + 1
• if high - low > best, then
• best := high - low
• ret := nums[high - 1]
• return ret

Let us see the following implementation to get better understanding:

## Example Code

Live Demo

class Solution:
def solve(self, nums, k):
nums.sort()
low, high = 0, 0
dist = 0
best = 0
ret = -1
while high < len(nums):
if high > 0 and nums[high] != nums[high - 1]:
dist += (high - low) * (nums[high] - nums[high - 1])
high += 1
while dist > k:
dist -= nums[high - 1] - nums[low]
low += 1
if high - low > best:
best = high - low
ret = nums[high - 1]
return ret

ob = Solution()
nums = [1, 0, 0, 0, 8, 8, 8, 8]
k = 8
print(ob.solve(nums, k))

## Input

[1, 0, 0, 0, 8, 8, 8, 8], 8

## Output

8