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2605. Form Smallest Number From Two Digit Arrays

Description

Given two arrays of unique digits nums1 and nums2, return the smallest number that contains at least one digit from each array.

 

Example 1:

Input: nums1 = [4,1,3], nums2 = [5,7]
Output: 15
Explanation: The number 15 contains the digit 1 from nums1 and the digit 5 from nums2. It can be proven that 15 is the smallest number we can have.

Example 2:

Input: nums1 = [3,5,2,6], nums2 = [3,1,7]
Output: 3
Explanation: The number 3 contains the digit 3 which exists in both arrays.

 

Constraints:

• 1 <= nums1.length, nums2.length <= 9
• 1 <= nums1[i], nums2[i] <= 9
• All digits in each array are unique.

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int minNumber(int[] nums1, int[] nums2) {
          int mask1 = 0, mask2 = 0;
          for (int x : nums1) {
              mask1 |= 1 << x;
          }
          for (int x : nums2) {
              mask2 |= 1 << x;
          }
          int mask = mask1 & mask2;
          if (mask != 0) {
              return Integer.numberOfTrailingZeros(mask);
          }
          int a = Integer.numberOfTrailingZeros(mask1);
          int b = Integer.numberOfTrailingZeros(mask2);
          return Math.min(a * 10 + b, b * 10 + a);
      }
  }
  

• class Solution {
  public:
      int minNumber(vector<int>& nums1, vector<int>& nums2) {
          int mask1 = 0, mask2 = 0;
          for (int x : nums1) {
              mask1 |= 1 << x;
          }
          for (int x : nums2) {
              mask2 |= 1 << x;
          }
          int mask = mask1 & mask2;
          if (mask) {
              return __builtin_ctz(mask);
          }
          int a = __builtin_ctz(mask1);
          int b = __builtin_ctz(mask2);
          return min(a * 10 + b, b * 10 + a);
      }
  };
  

• class Solution:
      def minNumber(self, nums1: List[int], nums2: List[int]) -> int:
          mask1 = mask2 = 0
          for x in nums1:
              mask1 |= 1 << x
          for x in nums2:
              mask2 |= 1 << x
          mask = mask1 & mask2
          if mask:
              return (mask & -mask).bit_length() - 1
          a = (mask1 & -mask1).bit_length() - 1
          b = (mask2 & -mask2).bit_length() - 1
          return min(a * 10 + b, b * 10 + a)
  
  

• func minNumber(nums1 []int, nums2 []int) int {
  	var mask1, mask2 uint
  	for _, x := range nums1 {
  		mask1 |= 1 << x
  	}
  	for _, x := range nums2 {
  		mask2 |= 1 << x
  	}
  	if mask := mask1 & mask2; mask != 0 {
  		return bits.TrailingZeros(mask)
  	}
  	a, b := bits.TrailingZeros(mask1), bits.TrailingZeros(mask2)
  	return min(a*10+b, b*10+a)
  }
  
  func min(a, b int) int {
  	if a < b {
  		return a
  	}
  	return b
  }
  

• function minNumber(nums1: number[], nums2: number[]): number {
      let mask1: number = 0;
      let mask2: number = 0;
      for (const x of nums1) {
          mask1 |= 1 << x;
      }
      for (const x of nums2) {
          mask2 |= 1 << x;
      }
      const mask = mask1 & mask2;
      if (mask !== 0) {
          return numberOfTrailingZeros(mask);
      }
      const a = numberOfTrailingZeros(mask1);
      const b = numberOfTrailingZeros(mask2);
      return Math.min(a * 10 + b, b * 10 + a);
  }
  
  function numberOfTrailingZeros(i: number): number {
      let y = 0;
      if (i === 0) {
          return 32;
      }
      let n = 31;
      y = i << 16;
      if (y != 0) {
          n = n - 16;
          i = y;
      }
      y = i << 8;
      if (y != 0) {
          n = n - 8;
          i = y;
      }
      y = i << 4;
      if (y != 0) {
          n = n - 4;
          i = y;
      }
      y = i << 2;
      if (y != 0) {
          n = n - 2;
          i = y;
      }
      return n - ((i << 1) >>> 31);
  }
  
  

• use std::collections::HashMap;
  
  impl Solution {
      pub fn min_number(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
          let mut h1: HashMap<i32, bool> = HashMap::new();
  
          for &n in &nums1 {
              h1.insert(n, true);
          }
  
          let mut h2: HashMap<i32, bool> = HashMap::new();
          for &n in &nums2 {
              h2.insert(n, true);
          }
  
          let mut a = 0;
          let mut b = 0;
          for i in 1..10 {
              if h1.contains_key(&i) && h2.contains_key(&i) {
                  return i;
              }
  
              if a == 0 && h1.contains_key(&i) {
                  a = i;
              }
  
              if b == 0 && h2.contains_key(&i) {
                  b = i;
              }
          }
  
  
          std::cmp::min(a * 10 + b, b * 10 + a)
      }
  }
  

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