Question

465. Optimal Account Balancing

中文文档

Description

You are given an array of transactions transactions where transactions[i] = [fromi, toi, amounti] indicates that the person with ID = fromi gave amounti $ to the person with ID = toi.

Return _the minimum number of transactions required to settle the debt_.

 

Example 1:

Input: transactions = [[0,1,10],[2,0,5]]
Output: 2
Explanation:
Person #0 gave person #1 $10.
Person #2 gave person #0 $5.
Two transactions are needed. One way to settle the debt is person #1 pays person #0 and #2 $5 each.

Example 2:

Input: transactions = [[0,1,10],[1,0,1],[1,2,5],[2,0,5]]
Output: 1
Explanation:
Person #0 gave person #1 $10.
Person #1 gave person #0 $1.
Person #1 gave person #2 $5.
Person #2 gave person #0 $5.
Therefore, person #1 only need to give person #0 $4, and all debt is settled.

 

Constraints:

• 1 <= transactions.length <= 8
• transactions[i].length == 3
• 0 <= fromi, toi < 12
• fromi != toi
• 1 <= amounti <= 100

Solutions

Solution 1

class Solution:
  def minTransfers(self, transactions: List[List[int]]) -> int:
    g = defaultdict(int)
    for f, t, x in transactions:
      g[f] -= x
      g[t] += x
    nums = [x for x in g.values() if x]
    m = len(nums)
    f = [inf] * (1 << m)
    f[0] = 0
    for i in range(1, 1 << m):
      s = 0
      for j, x in enumerate(nums):
        if i >> j & 1:
          s += x
      if s == 0:
        f[i] = i.bit_count() - 1
        j = (i - 1) & i
        while j > 0:
          f[i] = min(f[i], f[j] + f[i ^ j])
          j = (j - 1) & i
    return f[-1]

class Solution {
  public int minTransfers(int[][] transactions) {
    int[] g = new int[12];
    for (var t : transactions) {
      g[t[0]] -= t[2];
      g[t[1]] += t[2];
    }
    List<Integer> nums = new ArrayList<>();
    for (int x : g) {
      if (x != 0) {
        nums.add(x);
      }
    }
    int m = nums.size();
    int[] f = new int[1 << m];
    Arrays.fill(f, 1 << 29);
    f[0] = 0;
    for (int i = 1; i < 1 << m; ++i) {
      int s = 0;
      for (int j = 0; j < m; ++j) {
        if ((i >> j & 1) == 1) {
          s += nums.get(j);
        }
      }
      if (s == 0) {
        f[i] = Integer.bitCount(i) - 1;
        for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
          f[i] = Math.min(f[i], f[j] + f[i ^ j]);
        }
      }
    }
    return f[(1 << m) - 1];
  }
}

class Solution {
public:
  int minTransfers(vector<vector<int>>& transactions) {
    int g[12]{};
    for (auto& t : transactions) {
      g[t[0]] -= t[2];
      g[t[1]] += t[2];
    }
    vector<int> nums;
    for (int x : g) {
      if (x) {
        nums.push_back(x);
      }
    }
    int m = nums.size();
    int f[1 << m];
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for (int i = 1; i < 1 << m; ++i) {
      int s = 0;
      for (int j = 0; j < m; ++j) {
        if (i >> j & 1) {
          s += nums[j];
        }
      }
      if (s == 0) {
        f[i] = __builtin_popcount(i) - 1;
        for (int j = (i - 1) & i; j; j = (j - 1) & i) {
          f[i] = min(f[i], f[j] + f[i ^ j]);
        }
      }
    }
    return f[(1 << m) - 1];
  }
};

func minTransfers(transactions [][]int) int {
  g := [12]int{}
  for _, t := range transactions {
    g[t[0]] -= t[2]
    g[t[1]] += t[2]
  }
  nums := []int{}
  for _, x := range g {
    if x != 0 {
      nums = append(nums, x)
    }
  }
  m := len(nums)
  f := make([]int, 1<<m)
  for i := 1; i < 1<<m; i++ {
    f[i] = 1 << 29
    s := 0
    for j, x := range nums {
      if i>>j&1 == 1 {
        s += x
      }
    }
    if s == 0 {
      f[i] = bits.OnesCount(uint(i)) - 1
      for j := (i - 1) & i; j > 0; j = (j - 1) & i {
        f[i] = min(f[i], f[j]+f[i^j])
      }
    }
  }
  return f[1<<m-1]
}

function minTransfers(transactions: number[][]): number {
  const g: number[] = new Array(12).fill(0);
  for (const [f, t, x] of transactions) {
    g[f] -= x;
    g[t] += x;
  }
  const nums = g.filter(x => x !== 0);
  const m = nums.length;
  const f: number[] = new Array(1 << m).fill(1 << 29);
  f[0] = 0;
  for (let i = 1; i < 1 << m; ++i) {
    let s = 0;
    for (let j = 0; j < m; ++j) {
      if (((i >> j) & 1) === 1) {
        s += nums[j];
      }
    }
    if (s === 0) {
      f[i] = bitCount(i) - 1;
      for (let j = (i - 1) & i; j; j = (j - 1) & i) {
        f[i] = Math.min(f[i], f[j] + f[i ^ j]);
      }
    }
  }
  return f[(1 << m) - 1];
}

function bitCount(i: number): number {
  i = i - ((i >>> 1) & 0x55555555);
  i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
  i = (i + (i >>> 4)) & 0x0f0f0f0f;
  i = i + (i >>> 8);
  i = i + (i >>> 16);
  return i & 0x3f;
}