Question

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 ith 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 0th element one time. The cost is 2.
- Decrease the 1st element one time. The cost is 3.
- Decrease the 2nd 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.

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

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;
  }
}

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;
  }
};

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
}

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.

Similar problems:

• 296. Best Meeting Point
• 462. Minimum Moves to Equal Array Elements II

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)

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;
  }
}

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;
  }
};

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
}