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2448. Minimum Cost to Make Array Equal

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Description

You are given two 0-indexed arrays nums and cost consisting each of n positive integers.

You can do the following operation any number of times:

Increase or decrease any element of the array nums by 1.

The cost of doing one operation on the i^th element is cost[i].

Return the minimum total cost such that all the elements of the array nums become equal.

Example 1:

Input: nums = [1,3,5,2], cost = [2,3,1,14]
Output: 8
Explanation: We can make all the elements equal to 2 in the following way:
- Increase the 0^th element one time. The cost is 2.
- Decrease the 1^st element one time. The cost is 3.
- Decrease the 2^nd element three times. The cost is 1 + 1 + 1 = 3.
The total cost is 2 + 3 + 3 = 8.
It can be shown that we cannot make the array equal with a smaller cost.

Example 2:

Input: nums = [2,2,2,2,2], cost = [4,2,8,1,3]
Output: 0
Explanation: All the elements are already equal, so no operations are needed.

Constraints:

n == nums.length == cost.length
1 <= n <= 10⁵
1 <= nums[i], cost[i] <= 10⁶
Test cases are generated in a way that the output doesn't exceed 2⁵³-1

Solutions

Solution 1: Prefix Sum + Sorting + Enumeration

Let's denote the elements of the array nums as $a_1, a_2, \cdots, a_n$ and the elements of the array cost as $b_1, b_2, \cdots, b_n$. We can assume that $a_1 \leq a_2 \leq \cdots \leq a_n$, i.e., the array nums is sorted in ascending order.

Suppose we change all elements in the array nums to $x$, then the total cost we need is:

$$ \begin{aligned} \sum_{i=1}^{n} \left | a_i-x \right | b_i &= \sum_{i=1}^{k} (x-a_i)b_i + \sum_{i=k+1}^{n} (a_i-x)b_i \\ &= x\sum_{i=1}^{k} b_i - \sum_{i=1}^{k} a_ib_i + \sum_{i=k+1}^{n}a_ib_i - x\sum_{i=k+1}^{n}b_i \end{aligned} $$

where $k$ is the number of elements in $a_1, a_2, \cdots, a_n$ that are less than or equal to $x$.

We can use the prefix sum method to calculate $\sum_{i=1}^{k} b_i$ and $\sum_{i=1}^{k} a_ib_i$, as well as $\sum_{i=k+1}^{n}a_ib_i$ and $\sum_{i=k+1}^{n}b_i$.

Then we enumerate $x$, calculate the above four prefix sums, get the total cost mentioned above, and take the minimum value.

The time complexity is $O(n\times \log n)$, where $n$ is the length of the array nums. The main time complexity comes from sorting.

Python3

class Solution:
    def minCost(self, nums: List[int], cost: List[int]) -> int:
        arr = sorted(zip(nums, cost))
        n = len(arr)
        f = [0] * (n + 1)
        g = [0] * (n + 1)
        for i in range(1, n + 1):
            a, b = arr[i - 1]
            f[i] = f[i - 1] + a * b
            g[i] = g[i - 1] + b
        ans = inf
        for i in range(1, n + 1):
            a = arr[i - 1][0]
            l = a * g[i - 1] - f[i - 1]
            r = f[n] - f[i] - a * (g[n] - g[i])
            ans = min(ans, l + r)
        return ans

Java

class Solution {
    public long minCost(int[] nums, int[] cost) {
        int n = nums.length;
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; ++i) {
            arr[i] = new int[] {nums[i], cost[i]};
        }
        Arrays.sort(arr, (a, b) -> a[0] - b[0]);
        long[] f = new long[n + 1];
        long[] g = new long[n + 1];
        for (int i = 1; i <= n; ++i) {
            long a = arr[i - 1][0], b = arr[i - 1][1];
            f[i] = f[i - 1] + a * b;
            g[i] = g[i - 1] + b;
        }
        long ans = Long.MAX_VALUE;
        for (int i = 1; i <= n; ++i) {
            long a = arr[i - 1][0];
            long l = a * g[i - 1] - f[i - 1];
            long r = f[n] - f[i] - a * (g[n] - g[i]);
            ans = Math.min(ans, l + r);
        }
        return ans;
    }
}

C++

using ll = long long;

class Solution {
public:
    long long minCost(vector<int>& nums, vector<int>& cost) {
        int n = nums.size();
        vector<pair<int, int>> arr(n);
        for (int i = 0; i < n; ++i) arr[i] = {nums[i], cost[i]};
        sort(arr.begin(), arr.end());
        vector<ll> f(n + 1), g(n + 1);
        for (int i = 1; i <= n; ++i) {
            auto [a, b] = arr[i - 1];
            f[i] = f[i - 1] + 1ll * a * b;
            g[i] = g[i - 1] + b;
        }
        ll ans = 1e18;
        for (int i = 1; i <= n; ++i) {
            auto [a, _] = arr[i - 1];
            ll l = 1ll * a * g[i - 1] - f[i - 1];
            ll r = f[n] - f[i] - 1ll * a * (g[n] - g[i]);
            ans = min(ans, l + r);
        }
        return ans;
    }
};

Go

func minCost(nums []int, cost []int) int64 {
	n := len(nums)
	type pair struct{ a, b int }
	arr := make([]pair, n)
	for i, a := range nums {
		b := cost[i]
		arr[i] = pair{a, b}
	}
	sort.Slice(arr, func(i, j int) bool { return arr[i].a < arr[j].a })
	f := make([]int, n+1)
	g := make([]int, n+1)
	for i := 1; i <= n; i++ {
		a, b := arr[i-1].a, arr[i-1].b
		f[i] = f[i-1] + a*b
		g[i] = g[i-1] + b
	}
	var ans int64 = 1e18
	for i := 1; i <= n; i++ {
		a := arr[i-1].a
		l := a*g[i-1] - f[i-1]
		r := f[n] - f[i] - a*(g[n]-g[i])
		ans = min(ans, int64(l+r))
	}
	return ans
}

Rust

impl Solution {
    #[allow(dead_code)]
    pub fn min_cost(nums: Vec<i32>, cost: Vec<i32>) -> i64 {
        let mut zip_vec: Vec<_> = nums.into_iter().zip(cost.into_iter()).collect();

        // Sort the zip vector based on nums
        zip_vec.sort_by(|lhs, rhs| lhs.0.cmp(&rhs.0));

        let (nums, cost): (Vec<i32>, Vec<i32>) = zip_vec.into_iter().unzip();

        let mut sum: i64 = 0;
        for &c in &cost {
            sum += c as i64;
        }
        let middle_cost = (sum + 1) / 2;
        let mut cur_sum: i64 = 0;
        let mut i = 0;
        let n = nums.len();

        while i < n {
            if (cost[i] as i64) + cur_sum >= middle_cost {
                break;
            }
            cur_sum += cost[i] as i64;
            i += 1;
        }

        Self::compute_manhattan_dis(&nums, &cost, nums[i])
    }

    #[allow(dead_code)]
    fn compute_manhattan_dis(v: &Vec<i32>, c: &Vec<i32>, e: i32) -> i64 {
        let mut ret = 0;
        let n = v.len();

        for i in 0..n {
            if v[i] == e {
                continue;
            }
            ret += ((v[i] - e).abs() as i64) * (c[i] as i64);
        }

        ret
    }
}

Solution 2: Sorting + Median

We can also consider $b_i$ as the occurrence times of $a_i$, then the index of the median is $\frac{\sum_{i=1}^{n} b_i}{2}$. Changing all numbers to the median is definitely optimal.

The time complexity is $O(n\times \log n)$, where $n$ is the length of the array nums. The main time complexity comes from sorting.

Python3

class Solution:
    def minCost(self, nums: List[int], cost: List[int]) -> int:
        arr = sorted(zip(nums, cost))
        mid = sum(cost) // 2
        s = 0
        for x, c in arr:
            s += c
            if s > mid:
                return sum(abs(v - x) * c for v, c in arr)

Java

class Solution {
    public long minCost(int[] nums, int[] cost) {
        int n = nums.length;
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; ++i) {
            arr[i] = new int[] {nums[i], cost[i]};
        }
        Arrays.sort(arr, (a, b) -> a[0] - b[0]);
        long mid = sum(cost) / 2;
        long s = 0, ans = 0;
        for (var e : arr) {
            int x = e[0], c = e[1];
            s += c;
            if (s > mid) {
                for (var t : arr) {
                    ans += (long) Math.abs(t[0] - x) * t[1];
                }
                break;
            }
        }
        return ans;
    }

    private long sum(int[] arr) {
        long s = 0;
        for (int v : arr) {
            s += v;
        }
        return s;
    }
}

C++

using ll = long long;

class Solution {
public:
    long long minCost(vector<int>& nums, vector<int>& cost) {
        int n = nums.size();
        vector<pair<int, int>> arr(n);
        for (int i = 0; i < n; ++i) arr[i] = {nums[i], cost[i]};
        sort(arr.begin(), arr.end());
        ll mid = accumulate(cost.begin(), cost.end(), 0ll) / 2;
        ll s = 0, ans = 0;
        for (auto [x, c] : arr) {
            s += c;
            if (s > mid) {
                for (auto [v, d] : arr) {
                    ans += 1ll * abs(v - x) * d;
                }
                break;
            }
        }
        return ans;
    }
};

Go

func minCost(nums []int, cost []int) int64 {
	n := len(nums)
	type pair struct{ a, b int }
	arr := make([]pair, n)
	mid := 0
	for i, a := range nums {
		b := cost[i]
		mid += b
		arr[i] = pair{a, b}
	}
	mid /= 2
	sort.Slice(arr, func(i, j int) bool { return arr[i].a < arr[j].a })
	s, ans := 0, 0
	for _, e := range arr {
		x, c := e.a, e.b
		s += c
		if s > mid {
			for _, t := range arr {
				ans += abs(t.a-x) * t.b
			}
			break
		}
	}
	return int64(ans)

}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

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2448. Minimum Cost to Make Array Equal

Description

Solutions

Solution 1: Prefix Sum + Sorting + Enumeration

Python3

Java

C++

Go

Rust

Solution 2: Sorting + Median

Python3

Java

C++

Go

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README_EN.md

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2448. Minimum Cost to Make Array Equal

Description

Solutions

Solution 1: Prefix Sum + Sorting + Enumeration

Python3

Java

C++

Go

Rust

Solution 2: Sorting + Median

Python3

Java

C++

Go