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2035. Partition Array Into Two Arrays to Minimize Sum Difference
Description
You are given an integer array nums
of 2 * n
integers. You need to partition nums
into two arrays of length n
to minimize the absolute difference of the sums of the arrays. To partition nums
, put each element of nums
into one of the two arrays.
Return the minimum possible absolute difference.
Example 1:
Input: nums = [3,9,7,3] Output: 2 Explanation: One optimal partition is: [3,9] and [7,3]. The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.
Example 2:
Input: nums = [-36,36] Output: 72 Explanation: One optimal partition is: [-36] and [36]. The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.
Example 3:
Input: nums = [2,-1,0,4,-2,-9] Output: 0 Explanation: One optimal partition is: [2,4,-9] and [-1,0,-2]. The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.
Constraints:
1 <= n <= 15
nums.length == 2 * n
-107 <= nums[i] <= 107
Solutions
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class Solution { public int minimumDifference(int[] nums) { int n = nums.length >> 1; Map<Integer, Set<Integer>> f = new HashMap<>(); Map<Integer, Set<Integer>> g = new HashMap<>(); for (int i = 0; i < (1 << n); ++i) { int s = 0, cnt = 0; int s1 = 0, cnt1 = 0; for (int j = 0; j < n; ++j) { if ((i & (1 << j)) != 0) { s += nums[j]; ++cnt; s1 += nums[n + j]; ++cnt1; } else { s -= nums[j]; s1 -= nums[n + j]; } } f.computeIfAbsent(cnt, k -> new HashSet<>()).add(s); g.computeIfAbsent(cnt1, k -> new HashSet<>()).add(s1); } int ans = Integer.MAX_VALUE; for (int i = 0; i <= n; ++i) { List<Integer> fi = new ArrayList<>(f.get(i)); List<Integer> gi = new ArrayList<>(g.get(n - i)); Collections.sort(fi); Collections.sort(gi); for (int a : fi) { int left = 0, right = gi.size() - 1; int b = -a; while (left < right) { int mid = (left + right) >> 1; if (gi.get(mid) >= b) { right = mid; } else { left = mid + 1; } } ans = Math.min(ans, Math.abs(a + gi.get(left))); if (left > 0) { ans = Math.min(ans, Math.abs(a + gi.get(left - 1))); } } } return ans; } }
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class Solution { public: int minimumDifference(vector<int>& nums) { int n = nums.size() >> 1; vector<vector<int>> f(n + 1), g(n + 1); for (int i = 0; i < (1 << n); ++i) { int s = 0, cnt = 0; int s1 = 0, cnt1 = 0; for (int j = 0; j < n; ++j) { if (i & (1 << j)) { s += nums[j]; ++cnt; s1 += nums[n + j]; ++cnt1; } else { s -= nums[j]; s1 -= nums[n + j]; } } f[cnt].push_back(s); g[cnt1].push_back(s1); } for (int i = 0; i <= n; ++i) { sort(f[i].begin(), f[i].end()); sort(g[i].begin(), g[i].end()); } int ans = INT_MAX; for (int i = 0; i <= n; ++i) { for (int a : f[i]) { int left = 0, right = g[n - i].size() - 1; int b = -a; while (left < right) { int mid = (left + right) >> 1; if (g[n - i][mid] >= b) right = mid; else left = mid + 1; } ans = min(ans, abs(a + g[n - i][left])); if (left > 0) ans = min(ans, abs(a + g[n - i][left - 1])); } } return ans; } };
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class Solution: def minimumDifference(self, nums: List[int]) -> int: n = len(nums) >> 1 f = defaultdict(set) g = defaultdict(set) for i in range(1 << n): s = cnt = 0 s1 = cnt1 = 0 for j in range(n): if (i & (1 << j)) != 0: s += nums[j] cnt += 1 s1 += nums[n + j] cnt1 += 1 else: s -= nums[j] s1 -= nums[n + j] f[cnt].add(s) g[cnt1].add(s1) ans = inf for i in range(n + 1): fi, gi = sorted(list(f[i])), sorted(list(g[n - i])) # min(abs(f[i] + g[n - i])) for a in fi: left, right = 0, len(gi) - 1 b = -a while left < right: mid = (left + right) >> 1 if gi[mid] >= b: right = mid else: left = mid + 1 ans = min(ans, abs(a + gi[left])) if left > 0: ans = min(ans, abs(a + gi[left - 1])) return ans
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func minimumDifference(nums []int) int { n := len(nums) >> 1 f := make([][]int, n+1) g := make([][]int, n+1) for i := 0; i < (1 << n); i++ { s, cnt := 0, 0 s1, cnt1 := 0, 0 for j := 0; j < n; j++ { if (i & (1 << j)) != 0 { s += nums[j] cnt++ s1 += nums[n+j] cnt1++ } else { s -= nums[j] s1 -= nums[n+j] } } f[cnt] = append(f[cnt], s) g[cnt1] = append(g[cnt1], s1) } for i := 0; i <= n; i++ { sort.Ints(f[i]) sort.Ints(g[i]) } ans := 1 << 30 for i := 0; i <= n; i++ { for _, a := range f[i] { left, right := 0, len(g[n-i])-1 b := -a for left < right { mid := (left + right) >> 1 if g[n-i][mid] >= b { right = mid } else { left = mid + 1 } } ans = min(ans, abs(a+g[n-i][left])) if left > 0 { ans = min(ans, abs(a+g[n-i][left-1])) } } } return ans } func abs(x int) int { if x > 0 { return x } return -x }