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Medium
Bit Manipulation
Memoization
Array
Dynamic Programming
Backtracking
Bitmask

698. Partition to K Equal Sum Subsets

中文文档

Description

Given an integer array nums and an integer k, return true if it is possible to divide this array into k non-empty subsets whose sums are all equal.

 

Example 1:

Input: nums = [4,3,2,3,5,2,1], k = 4
Output: true
Explanation: It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.

Example 2:

Input: nums = [1,2,3,4], k = 3
Output: false

 

Constraints:

  • 1 <= k <= nums.length <= 16
  • 1 <= nums[i] <= 104
  • The frequency of each element is in the range [1, 4].

Solutions

Solution 1: DFS + Pruning

According to the problem description, we need to partition the array $\textit{nums}$ into $k$ subsets such that the sum of each subset is equal. Therefore, we first sum all the elements in $\textit{nums}$. If the total sum cannot be divided by $k$, it means we cannot partition the array into $k$ subsets, and we return $\textit{false}$ early.

If the total sum can be divided by $k$, let's denote the expected sum of each subset as $s$. Then, we create an array $\textit{cur}$ of length $k$ to represent the current sum of each subset.

We sort the array $\textit{nums}$ in descending order (to reduce the number of searches), and then start from the first element, trying to add it to each subset in $\textit{cur}$ one by one. If adding $\textit{nums}[i]$ to a subset $\textit{cur}[j]$ makes the subset's sum exceed $s$, it means it cannot be placed in that subset, and we can skip it. Additionally, if $\textit{cur}[j]$ is equal to $\textit{cur}[j - 1]$, it means we have already completed the search for $\textit{cur}[j - 1]$, and we can skip the current search.

If we can add all elements to $\textit{cur}$, it means we can partition the array into $k$ subsets, and we return $\textit{true}$.

Python3

class Solution:
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        def dfs(i: int) -> bool:
            if i == len(nums):
                return True
            for j in range(k):
                if j and cur[j] == cur[j - 1]:
                    continue
                cur[j] += nums[i]
                if cur[j] <= s and dfs(i + 1):
                    return True
                cur[j] -= nums[i]
            return False

        s, mod = divmod(sum(nums), k)
        if mod:
            return False
        cur = [0] * k
        nums.sort(reverse=True)
        return dfs(0)

Java

class Solution {
    private int[] nums;
    private int[] cur;
    private int s;

    public boolean canPartitionKSubsets(int[] nums, int k) {
        for (int v : nums) {
            s += v;
        }
        if (s % k != 0) {
            return false;
        }
        s /= k;
        cur = new int[k];
        Arrays.sort(nums);
        this.nums = nums;
        return dfs(nums.length - 1);
    }

    private boolean dfs(int i) {
        if (i < 0) {
            return true;
        }
        for (int j = 0; j < cur.length; ++j) {
            if (j > 0 && cur[j] == cur[j - 1]) {
                continue;
            }
            cur[j] += nums[i];
            if (cur[j] <= s && dfs(i - 1)) {
                return true;
            }
            cur[j] -= nums[i];
        }
        return false;
    }
}

C++

class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s % k) {
            return false;
        }
        s /= k;
        int n = nums.size();
        vector<int> cur(k);
        function<bool(int)> dfs = [&](int i) {
            if (i == n) {
                return true;
            }
            for (int j = 0; j < k; ++j) {
                if (j && cur[j] == cur[j - 1]) {
                    continue;
                }
                cur[j] += nums[i];
                if (cur[j] <= s && dfs(i + 1)) {
                    return true;
                }
                cur[j] -= nums[i];
            }
            return false;
        };
        sort(nums.begin(), nums.end(), greater<int>());
        return dfs(0);
    }
};

Go

func canPartitionKSubsets(nums []int, k int) bool {
	s := 0
	for _, v := range nums {
		s += v
	}
	if s%k != 0 {
		return false
	}
	s /= k
	cur := make([]int, k)
	n := len(nums)

	var dfs func(int) bool
	dfs = func(i int) bool {
		if i == n {
			return true
		}
		for j := 0; j < k; j++ {
			if j > 0 && cur[j] == cur[j-1] {
				continue
			}
			cur[j] += nums[i]
			if cur[j] <= s && dfs(i+1) {
				return true
			}
			cur[j] -= nums[i]
		}
		return false
	}

	sort.Sort(sort.Reverse(sort.IntSlice(nums)))
	return dfs(0)
}

TypeScript

function canPartitionKSubsets(nums: number[], k: number): boolean {
    const dfs = (i: number): boolean => {
        if (i === nums.length) {
            return true;
        }
        for (let j = 0; j < k; j++) {
            if (j > 0 && cur[j] === cur[j - 1]) {
                continue;
            }
            cur[j] += nums[i];
            if (cur[j] <= s && dfs(i + 1)) {
                return true;
            }
            cur[j] -= nums[i];
        }
        return false;
    };

    let s = nums.reduce((a, b) => a + b, 0);
    const mod = s % k;
    if (mod !== 0) {
        return false;
    }
    s = Math.floor(s / k);
    const cur = Array(k).fill(0);
    nums.sort((a, b) => b - a);
    return dfs(0);
}

Solution 2: State Compression + Memoization

Similar to Solution 1, we first check whether the array $\textit{nums}$ can be partitioned into $k$ subsets. If it cannot be divided by $k$, we directly return $\textit{false}$.

Let $s$ be the expected sum of each subset, and let $\textit{state}$ represent the current partitioning state of the elements. For the $i$-th number, if the $i$-th bit of $\textit{state}$ is $0$, it means the $i$-th element has not been partitioned.

Our goal is to form $k$ subsets with a sum of $s$ from all elements. Let $t$ be the current sum of the subset. When the $i$-th element is not partitioned:

  • If $t + \textit{nums}[i] \gt s$, it means the $i$-th element cannot be added to the current subset. Since we sort the array $\textit{nums}$ in ascending order, all elements from position $i$ onwards cannot be added to the current subset, and we directly return $\textit{false}$.
  • Otherwise, add the $i$-th element to the current subset, change the state to $\textit{state} | 2^i$, and continue searching for unpartitioned elements. Note that if $t + \textit{nums}[i] = s$, it means we can form a subset with a sum of $s$. The next step is to reset $t$ to zero (which can be achieved by $(t + \textit{nums}[i]) \bmod s$) and continue partitioning the next subset.

To avoid repeated searches, we use an array $\textit{f}$ of length $2^n$ to record the search results for each state. The array $\textit{f}$ has three possible values:

  • 0 indicates that the current state has not been searched yet;
  • -1: indicates that the current state cannot be partitioned into $k$ subsets;
  • 1: indicates that the current state can be partitioned into $k$ subsets.

The time complexity is $O(n \times 2^n)$, and the space complexity is $O(2^n)$. Here, $n$ represents the length of the array $\textit{nums}$. For each state, we need to traverse the array $\textit{nums}$, which has a time complexity of $O(n)$. The total number of states is $2^n$, so the overall time complexity is $O(n \times 2^n)$.

Python3

class Solution:
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        @cache
        def dfs(state, t):
            if state == mask:
                return True
            for i, v in enumerate(nums):
                if (state >> i) & 1:
                    continue
                if t + v > s:
                    break
                if dfs(state | 1 << i, (t + v) % s):
                    return True
            return False

        s, mod = divmod(sum(nums), k)
        if mod:
            return False
        nums.sort()
        mask = (1 << len(nums)) - 1
        return dfs(0, 0)

Java

class Solution {
    private int[] f;
    private int[] nums;
    private int n;
    private int s;

    public boolean canPartitionKSubsets(int[] nums, int k) {
        for (int v : nums) {
            s += v;
        }
        if (s % k != 0) {
            return false;
        }
        s /= k;
        Arrays.sort(nums);
        this.nums = nums;
        n = nums.length;
        f = new int[1 << n];
        return dfs(0, 0);
    }

    private boolean dfs(int state, int t) {
        if (state == (1 << n) - 1) {
            return true;
        }
        if (f[state] != 0) {
            return f[state] == 1;
        }
        for (int i = 0; i < n; ++i) {
            if (((state >> i) & 1) == 1) {
                continue;
            }
            if (t + nums[i] > s) {
                break;
            }
            if (dfs(state | 1 << i, (t + nums[i]) % s)) {
                f[state] = 1;
                return true;
            }
        }
        f[state] = -1;
        return false;
    }
}

C++

class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s % k) {
            return false;
        }
        s /= k;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        int mask = (1 << n) - 1;
        vector<int> f(1 << n);
        function<bool(int, int)> dfs = [&](int state, int t) {
            if (state == mask) {
                return true;
            }
            if (f[state]) {
                return f[state] == 1;
            }
            for (int i = 0; i < n; ++i) {
                if (state >> i & 1) {
                    continue;
                }
                if (t + nums[i] > s) {
                    break;
                }
                if (dfs(state | 1 << i, (t + nums[i]) % s)) {
                    f[state] = 1;
                    return true;
                }
            }
            f[state] = -1;
            return false;
        };
        return dfs(0, 0);
    }
};

Go

func canPartitionKSubsets(nums []int, k int) bool {
	s := 0
	for _, v := range nums {
		s += v
	}
	if s%k != 0 {
		return false
	}
	s /= k
	n := len(nums)
	f := make([]int, 1<<n)
	mask := (1 << n) - 1

	var dfs func(int, int) bool
	dfs = func(state, t int) bool {
		if state == mask {
			return true
		}
		if f[state] != 0 {
			return f[state] == 1
		}
		for i, v := range nums {
			if (state >> i & 1) == 1 {
				continue
			}
			if t+v > s {
				break
			}
			if dfs(state|1<<i, (t+v)%s) {
				f[state] = 1
				return true
			}
		}
		f[state] = -1
		return false
	}

	sort.Ints(nums)
	return dfs(0, 0)
}

TypeScript

function canPartitionKSubsets(nums: number[], k: number): boolean {
    let s = nums.reduce((a, b) => a + b, 0);
    if (s % k !== 0) {
        return false;
    }
    s = Math.floor(s / k);
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const mask = (1 << n) - 1;
    const f = Array(1 << n).fill(0);

    const dfs = (state: number, t: number): boolean => {
        if (state === mask) {
            return true;
        }
        if (f[state] !== 0) {
            return f[state] === 1;
        }
        for (let i = 0; i < n; ++i) {
            if ((state >> i) & 1) {
                continue;
            }
            if (t + nums[i] > s) {
                break;
            }
            if (dfs(state | (1 << i), (t + nums[i]) % s)) {
                f[state] = 1;
                return true;
            }
        }
        f[state] = -1;
        return false;
    };

    return dfs(0, 0);
}

Solution 3: Dynamic Programming

We can use dynamic programming to solve this problem.

We define $f[i]$ to represent whether there exists $k$ subsets that meet the requirements when the current selected numbers' state is $i$. Initially, $f[0] = \text{true}$, and the answer is $f[2^n - 1]$, where $n$ is the length of the array $\textit{nums}$. Additionally, we define $cur[i]$ to represent the sum of the last subset when the current selected numbers' state is $i$.

We enumerate the states $i$ in the range $[0, 2^n]$. For each state $i$, if $f[i]$ is $\text{false}$, we skip it. Otherwise, we enumerate any number $\textit{nums}[j]$ in the array $\textit{nums}$. If $cur[i] + \textit{nums}[j] &gt; s$, we break the enumeration loop because the subsequent numbers are larger and cannot be placed in the current subset. Otherwise, if the $j$-th bit of the binary representation of $i$ is $0$, it means the current $\textit{nums}[j]$ has not been selected. We can place it in the current subset, change the state to $i | 2^j$, update $cur[i | 2^j] = (cur[i] + \textit{nums}[j]) \bmod s$, and set $f[i | 2^j] = \text{true}$.

Finally, we return $f[2^n - 1]$.

The time complexity is $O(n \times 2^n)$, and the space complexity is $O(2^n)$. Here, $n$ represents the length of the array $\textit{nums}$.

Python3

class Solution:
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        s = sum(nums)
        if s % k:
            return False
        s //= k
        nums.sort()
        n = len(nums)
        f = [False] * (1 << n)
        cur = [0] * (1 << n)
        f[0] = True
        for i in range(1 << n):
            if not f[i]:
                continue
            for j in range(n):
                if cur[i] + nums[j] > s:
                    break
                if (i >> j & 1) == 0:
                    if not f[i | 1 << j]:
                        cur[i | 1 << j] = (cur[i] + nums[j]) % s
                        f[i | 1 << j] = True
        return f[-1]

Java

class Solution {
    public boolean canPartitionKSubsets(int[] nums, int k) {
        int s = 0;
        for (int x : nums) {
            s += x;
        }
        if (s % k != 0) {
            return false;
        }
        s /= k;
        Arrays.sort(nums);
        int n = nums.length;
        boolean[] f = new boolean[1 << n];
        f[0] = true;
        int[] cur = new int[1 << n];
        for (int i = 0; i < 1 << n; ++i) {
            if (!f[i]) {
                continue;
            }
            for (int j = 0; j < n; ++j) {
                if (cur[i] + nums[j] > s) {
                    break;
                }
                if ((i >> j & 1) == 0) {
                    cur[i | 1 << j] = (cur[i] + nums[j]) % s;
                    f[i | 1 << j] = true;
                }
            }
        }
        return f[(1 << n) - 1];
    }
}

C++

class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s % k) {
            return false;
        }
        s /= k;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        bool f[1 << n];
        int cur[1 << n];
        memset(f, false, sizeof(f));
        memset(cur, 0, sizeof(cur));
        f[0] = 1;
        for (int i = 0; i < 1 << n; ++i) {
            if (!f[i]) {
                continue;
            }
            for (int j = 0; j < n; ++j) {
                if (cur[i] + nums[j] > s) {
                    break;
                }
                if ((i >> j & 1) == 0) {
                    f[i | 1 << j] = true;
                    cur[i | 1 << j] = (cur[i] + nums[j]) % s;
                }
            }
        }
        return f[(1 << n) - 1];
    }
};

Go

func canPartitionKSubsets(nums []int, k int) bool {
	s := 0
	for _, x := range nums {
		s += x
	}
	if s%k != 0 {
		return false
	}
	s /= k
	sort.Ints(nums)
	n := len(nums)
	f := make([]bool, 1<<n)
	cur := make([]int, 1<<n)
	f[0] = true
	for i := 0; i < 1<<n; i++ {
		if !f[i] {
			continue
		}
		for j := 0; j < n; j++ {
			if cur[i]+nums[j] > s {
				break
			}
			if i>>j&1 == 0 {
				f[i|1<<j] = true
				cur[i|1<<j] = (cur[i] + nums[j]) % s
			}
		}
	}
	return f[(1<<n)-1]
}

TypeScript

function canPartitionKSubsets(nums: number[], k: number): boolean {
    let s = nums.reduce((a, b) => a + b);
    if (s % k !== 0) {
        return false;
    }
    s /= k;
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const f: boolean[] = Array(1 << n).fill(false);
    f[0] = true;
    const cur: number[] = Array(n).fill(0);
    for (let i = 0; i < 1 << n; ++i) {
        if (!f[i]) {
            continue;
        }
        for (let j = 0; j < n; ++j) {
            if (cur[i] + nums[j] > s) {
                break;
            }
            if (((i >> j) & 1) === 0) {
                f[i | (1 << j)] = true;
                cur[i | (1 << j)] = (cur[i] + nums[j]) % s;
            }
        }
    }
    return f[(1 << n) - 1];
}