# 2602. Minimum Operations to Make All Array Elements Equal

## Description

You are given an array nums consisting of positive integers.

You are also given an integer array queries of size m. For the ith query, you want to make all of the elements of nums equal to queries[i]. You can perform the following operation on the array any number of times:

• Increase or decrease an element of the array by 1.

Return an array answer of size m where answer[i] is the minimum number of operations to make all elements of nums equal to queries[i].

Note that after each query the array is reset to its original state.

Example 1:

Input: nums = [3,1,6,8], queries = [1,5]
Output: [14,10]
Explanation: For the first query we can do the following operations:
- Decrease nums[0] 2 times, so that nums = [1,1,6,8].
- Decrease nums[2] 5 times, so that nums = [1,1,1,8].
- Decrease nums[3] 7 times, so that nums = [1,1,1,1].
So the total number of operations for the first query is 2 + 5 + 7 = 14.
For the second query we can do the following operations:
- Increase nums[0] 2 times, so that nums = [5,1,6,8].
- Increase nums[1] 4 times, so that nums = [5,5,6,8].
- Decrease nums[2] 1 time, so that nums = [5,5,5,8].
- Decrease nums[3] 3 times, so that nums = [5,5,5,5].
So the total number of operations for the second query is 2 + 4 + 1 + 3 = 10.


Example 2:

Input: nums = [2,9,6,3], queries = [10]
Output: [20]
Explanation: We can increase each value in the array to 10. The total number of operations will be 8 + 1 + 4 + 7 = 20.


Constraints:

• n == nums.length
• m == queries.length
• 1 <= n, m <= 105
• 1 <= nums[i], queries[i] <= 109

## Solutions

Solution 1: sort + prefix sum + binary search

First, we sort the array $nums$ and calculate the prefix sum array $s$ with a length of $n+1$, where $s[i]$ represents the sum of the first $i$ elements in the array $nums$.

Then, we traverse each query $queries[i]$, we need to reduce all elements greater than $queries[i]$ to $queries[i]$, and increase all elements less than $queries[i]$ to $queries[i]$.

We can use binary search to find the index $i$ of the first element in the array $nums$ that is greater than $queries[i]$. There are $n-i$ elements that need to be reduced to $queries[i]$, and the sum of these elements is $s[n]-s[i]$. These elements need to be reduced by $n-i$ $queries[i]$, so the total number of operations to reduce these elements to $queries[i]$ is $s[n]-s[i]-(n-i)\times queries[i]$.

Similarly, we can find the index $i$ of the first element in the array $nums$ that is greater than or equal to $queries[i]$. There are $i$ elements that need to be increased to $queries[i]$, and the sum of these elements is $s[i]$. Therefore, the total number of operations to increase these elements to $queries[i]$ is $queries[i]\times i-s[i]$.

Finally, add these two total operation counts together to get the minimum number of operations to change all elements in the array $nums$ to $queries[i]$, that is, $ans[i]=s[n]-s[i]-(n-i)\times queries[i]+queries[i]\times i-s[i]$.

Time complexity $O(n \times \log n)$, space complexity $O(n)$, where $n$ is the length of the array $nums$.

• class Solution {
public List<Long> minOperations(int[] nums, int[] queries) {
Arrays.sort(nums);
int n = nums.length;
long[] s = new long[n + 1];
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
List<Long> ans = new ArrayList<>();
for (int x : queries) {
int i = search(nums, x + 1);
long t = s[n] - s[i] - 1L * (n - i) * x;
i = search(nums, x);
t += 1L * x * i - s[i];
}
return ans;
}

private int search(int[] nums, int x) {
int l = 0, r = nums.length;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
}

• class Solution {
public:
vector<long long> minOperations(vector<int>& nums, vector<int>& queries) {
sort(nums.begin(), nums.end());
int n = nums.size();
vector<long long> s(n + 1);
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
vector<long long> ans;
for (auto& x : queries) {
int i = lower_bound(nums.begin(), nums.end(), x + 1) - nums.begin();
long long t = s[n] - s[i] - 1LL * (n - i) * x;
i = lower_bound(nums.begin(), nums.end(), x) - nums.begin();
t += 1LL * x * i - s[i];
ans.push_back(t);
}
return ans;
}
};

• class Solution:
def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:
nums.sort()
s = list(accumulate(nums, initial=0))
ans = []
for x in queries:
i = bisect_left(nums, x + 1)
t = s[-1] - s[i] - (len(nums) - i) * x
i = bisect_left(nums, x)
t += x * i - s[i]
ans.append(t)
return ans


• func minOperations(nums []int, queries []int) (ans []int64) {
sort.Ints(nums)
n := len(nums)
s := make([]int, n+1)
for i, x := range nums {
s[i+1] = s[i] + x
}
for _, x := range queries {
i := sort.SearchInts(nums, x+1)
t := s[n] - s[i] - (n-i)*x
i = sort.SearchInts(nums, x)
t += x*i - s[i]
ans = append(ans, int64(t))
}
return
}

• function minOperations(nums: number[], queries: number[]): number[] {
nums.sort((a, b) => a - b);
const n = nums.length;
const s: number[] = new Array(n + 1).fill(0);
for (let i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
const search = (x: number): number => {
let l = 0;
let r = n;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
const ans: number[] = [];
for (const x of queries) {
const i = search(x + 1);
let t = s[n] - s[i] - (n - i) * x;
const j = search(x);
t += x * j - s[j];
ans.push(t);
}
return ans;
}