# Solution For Minimum Average Difference

The Minimum Average Difference problem on LeetCode involves finding the minimum possible difference between the average of two groups of integers, given an array of integers. Here is a detailed solution to this problem:

Approach:

The brute force approach to solve this problem would involve taking all possible combinations of two sets and then taking their averages, computing the difference and then finding the minimum difference among all the combinations. However, the time complexity of this approach would be O(n^2 * 2^n), which is not efficient enough to pass the LeetCode test cases.

One better approach is to use dynamic programming to keep track of all possible sums that can be obtained from subsets of the array. This can be done using a 2D boolean array dp[i][j]. Here, dp[i][j] represents whether a sum j can be obtained using the first i elements of the array. The base case for this DP is dp[0][0] = true, as we can obtain a sum of 0 using an empty set.

Once we have computed all possible sums for subsets of the array, we can iterate over all possible sums half of the total sum of array to 0 and check if this sum can be obtained using the first n/2 elements of the array. If so, we compute the difference between this sum and the remaining half and update our answer if this difference is lesser than the current minimum difference.

Solution:

Here is the Python code for the minimum average difference problem on LeetCode:

```class Solution: def minimumAverageDifference(self, nums: List[int]) -> float: n = len(nums) total = sum(nums) dp = [[False]*(total+1) for _ in range(n+1)] dp[0][0] = True for i in range(1, n+1): for j in range(total+1): if j-nums[i-1] >= 0 and dp[i-1][j-nums[i-1]]: dp[i][j] = True if dp[i-1][j]: dp[i][j] = True ans = float('inf') for i in range(total//2, -1, -1): if dp[n][i]: ans = min(ans, total-2*i) return ans/2```

Time Complexity:

The time complexity of this solution is O(n * total), where total is the sum of the elements in nums, since we are iterating over n * total elements for the DP, and then iterating once more over values of i from total//2 to 0.

Space Complexity:

The space complexity of this solution is O(n * total), since we are using a 2D array of size (n+1) * (total+1) for the DP.

## Step by Step Implementation For Minimum Average Difference

```Given an array of integers A, find the minimum of (A[i]-i) for all 0 <= i < A.length.

If there are multiple answers, return any of them.

class Solution {
public int minAverageDifference(int[] A) {
// TODO: Write your code here
int n = A.length;
int minAvg = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
int currAvg = 0;
for (int j = 0; j < n; j++) {
currAvg += Math.abs(A[i] - A[j]);
}
minAvg = Math.min(minAvg, currAvg);
}
return minAvg;
}
}```
```Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total.  (We may choose the same index i multiple times.)

Return the largest possible sum of the array after modifying it in this way.

def largestSumAfterKNegations(A, K):

# First, we sort the array in ascending order
A.sort()

# Then, we iterate through the array
# For each element, we check if making it negative
# would give us a larger sum. If so, we do so.
# Otherwise, we leave it as is.
for i in range(len(A)):
if K > 0 and A[i] < 0:
A[i] = -A[i]
K -= 1

# Finally, we return the sum of the array
return sum(A)```
```/**
* @param {number[]} nums
* @return {number}
*/

//Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value.

//Solution:

var findMaxAverage = function(nums, k) {
let sum = 0;
let maxSum = 0;

for(let i=0; iOne solution to this problem could be to keep track of the minimum average difference seen so far while iterating through the given array of integers. Whenever a new integer is encountered, calculate the average of that integer with all the other integers seen so far and update the minimum average difference accordingly. The code for this solution could look something like this:

int minAverageDifference(vector &nums) {
int minDiff = INT_MAX;
for (int i = 0; i < nums.size(); i++) {
for (int j = 0; j < i; j++) {
int avg = (nums[i] + nums[j]) / 2;
int diff = abs(avg - nums[i]);
minDiff = min(minDiff, diff);
}
}
return minDiff;
}using System;

public class Solution {

public int MinimumAverageDifference(int[] nums) {