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Description

Given an integer array nums, return true if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or false otherwise.

 

Example 1:

Input: nums = [1,5,11,5]
Output: true
Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

Input: nums = [1,2,3,5]
Output: false
Explanation: The array cannot be partitioned into equal sum subsets.

 

Constraints:

  • 1 <= nums.length <= 200
  • 1 <= nums[i] <= 100

Solutions

Python3

class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        m, mod = divmod(sum(nums), 2)
        if mod:
            return False
        n = len(nums)
        f = [[False] * (m + 1) for _ in range(n + 1)]
        f[0][0] = True
        for i, x in enumerate(nums, 1):
            for j in range(m + 1):
                f[i][j] = f[i - 1][j] or (j >= x and f[i - 1][j - x])
        return f[n][m]
class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        m, mod = divmod(sum(nums), 2)
        if mod:
            return False
        f = [True] + [False] * m
        for x in nums:
            for j in range(m, x - 1, -1):
                f[j] = f[j] or f[j - x]
        return f[m]

Java

class Solution {
    public boolean canPartition(int[] nums) {
        // int s = Arrays.stream(nums).sum();
        int s = 0;
        for (int x : nums) {
            s += x;
        }
        if (s % 2 == 1) {
            return false;
        }
        int n = nums.length;
        int m = s >> 1;
        boolean[][] f = new boolean[n + 1][m + 1];
        f[0][0] = true;
        for (int i = 1; i <= n; ++i) {
            int x = nums[i - 1];
            for (int j = 0; j <= m; ++j) {
                f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
            }
        }
        return f[n][m];
    }
}
class Solution {
    public boolean canPartition(int[] nums) {
        // int s = Arrays.stream(nums).sum();
        int s = 0;
        for (int x : nums) {
            s += x;
        }
        if (s % 2 == 1) {
            return false;
        }
        int m = s >> 1;
        boolean[] f = new boolean[m + 1];
        f[0] = true;
        for (int x : nums) {
            for (int j = m; j >= x; --j) {
                f[j] |= f[j - x];
            }
        }
        return f[m];
    }
}

C++

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s % 2 == 1) {
            return false;
        }
        int n = nums.size();
        int m = s >> 1;
        bool f[n + 1][m + 1];
        memset(f, false, sizeof(f));
        f[0][0] = true;
        for (int i = 1; i <= n; ++i) {
            int x = nums[i - 1];
            for (int j = 0; j <= m; ++j) {
                f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
            }
        }
        return f[n][m];
    }
};
class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int s = accumulate(nums.begin(), nums.end(), 0);
        if (s % 2 == 1) {
            return false;
        }
        int m = s >> 1;
        bool f[m + 1];
        memset(f, false, sizeof(f));
        f[0] = true;
        for (int& x : nums) {
            for (int j = m; j >= x; --j) {
                f[j] |= f[j - x];
            }
        }
        return f[m];
    }
};

Rust

impl Solution {
    #[allow(dead_code)]
    pub fn can_partition(nums: Vec<i32>) -> bool {
        let mut sum = 0;
        for e in &nums {
            sum += *e;
        }

        if sum % 2 != 0 {
            return false;
        }

        let n = nums.len();
        let m = (sum / 2) as usize;
        let mut dp: Vec<Vec<bool>> = vec![vec![false; m + 1]; n + 1];

        // Initialize the dp vector
        dp[0][0] = true;

        // Begin the actual dp process
        for i in 1..=n {
            for j in 0..=m {
                dp[i][j] = if (nums[i - 1] as usize) > j {
                    dp[i - 1][j]
                } else {
                    dp[i - 1][j] || dp[i - 1][j - (nums[i - 1] as usize)]
                };
            }
        }

        dp[n][m]
    }
}
impl Solution {
    #[allow(dead_code)]
    pub fn can_partition(nums: Vec<i32>) -> bool {
        let mut sum = 0;
        for e in &nums {
            sum += *e;
        }

        if sum % 2 != 0 {
            return false;
        }

        let m = (sum >> 1) as usize;

        // Here dp[i] means if it can be sum up to `i` for all the number we've traversed through so far
        // Which is actually compressing the 2-D dp vector to 1-D
        let mut dp: Vec<bool> = vec![false; m + 1];

        // Initialize the dp vector
        dp[0] = true;

        // Begin the actual dp process
        for e in &nums {
            // For every num in nums vector
            for i in (*e as usize..=m).rev() {
                // Update the current status
                dp[i] |= dp[i - (*e as usize)];
            }
        }

        dp[m]
    }
}

Go

func canPartition(nums []int) bool {
	s := 0
	for _, x := range nums {
		s += x
	}
	if s%2 == 1 {
		return false
	}
	n, m := len(nums), s>>1
	f := make([][]bool, n+1)
	for i := range f {
		f[i] = make([]bool, m+1)
	}
	f[0][0] = true
	for i := 1; i <= n; i++ {
		x := nums[i-1]
		for j := 0; j <= m; j++ {
			f[i][j] = f[i-1][j] || (j >= x && f[i-1][j-x])
		}
	}
	return f[n][m]
}
func canPartition(nums []int) bool {
	s := 0
	for _, x := range nums {
		s += x
	}
	if s%2 == 1 {
		return false
	}
	m := s >> 1
	f := make([]bool, m+1)
	f[0] = true
	for _, x := range nums {
		for j := m; j >= x; j-- {
			f[j] = f[j] || f[j-x]
		}
	}
	return f[m]
}

TypeScript

function canPartition(nums: number[]): boolean {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s % 2 === 1) {
        return false;
    }
    const n = nums.length;
    const m = s >> 1;
    const f: boolean[][] = Array(n + 1)
        .fill(0)
        .map(() => Array(m + 1).fill(false));
    f[0][0] = true;
    for (let i = 1; i <= n; ++i) {
        const x = nums[i - 1];
        for (let j = 0; j <= m; ++j) {
            f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
        }
    }
    return f[n][m];
}
function canPartition(nums: number[]): boolean {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s % 2 === 1) {
        return false;
    }
    const m = s >> 1;
    const f: boolean[] = Array(m + 1).fill(false);
    f[0] = true;
    for (const x of nums) {
        for (let j = m; j >= x; --j) {
            f[j] = f[j] || f[j - x];
        }
    }
    return f[m];
}

JavaScript

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var canPartition = function (nums) {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s % 2 === 1) {
        return false;
    }
    const n = nums.length;
    const m = s >> 1;
    const f = Array(n + 1)
        .fill(0)
        .map(() => Array(m + 1).fill(false));
    f[0][0] = true;
    for (let i = 1; i <= n; ++i) {
        const x = nums[i - 1];
        for (let j = 0; j <= m; ++j) {
            f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
        }
    }
    return f[n][m];
};
/**
 * @param {number[]} nums
 * @return {boolean}
 */
var canPartition = function (nums) {
    const s = nums.reduce((a, b) => a + b, 0);
    if (s % 2 === 1) {
        return false;
    }
    const m = s >> 1;
    const f = Array(m + 1).fill(false);
    f[0] = true;
    for (const x of nums) {
        for (let j = m; j >= x; --j) {
            f[j] = f[j] || f[j - x];
        }
    }
    return f[m];
};

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