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3897. Maximum Value of Concatenated Binary Segments
Hard

You are given two integer arrays nums1 and nums0, each of size n.

  • nums1[i] represents the number of '1's in the ith segment.
  • nums0[i] represents the number of '0's in the ith segment.

For each index i, construct a binary segment consisting of:

  • nums1[i] occurrences of '1' followed by
  • nums0[i] occurrences of '0'.

You may rearrange the order of these segments in any way. After rearranging, concatenate all segments to form a single binary string.

Return the maximum possible integer value of the concatenated binary string.

Since the result can be very large, return the answer modulo 109 + 7.

 

Example 1:

Input: nums1 = [1,2], nums0 = [1,0]

Output: 14

Explanation:

  • At index 0, nums1[0] = 1 and nums0[0] = 1, so the segment formed is "10".
  • At index 1, nums1[1] = 2 and nums0[1] = 0, so the segment formed is "11".
  • Reordering the segments as "11" followed by "10" produces the binary string "1110".
  • The binary number "1110" has value 14 which is the maximum possible value.

Example 2:

Input: nums1 = [3,1], nums0 = [0,3]

Output: 120

Explanation:

  • At index 0, nums1[0] = 3 and nums0[0] = 0, so the segment formed is "111".
  • At index 1, nums1[1] = 1 and nums0[1] = 3, so the segment formed is "1000".
  • Reordering the segments as "111" followed by "1000" produces the binary string "1111000".
  • The binary number "1111000" has value 120 which is the maximum possible value.

 

Constraints:

  • 1 <= n == nums1.length == nums0.length <= 105
  • 0 <= nums1[i], nums0[i] <= 104
  • nums1[i] + nums0[i] > 0
  • The total sum of all elements in nums1 and nums0 does not exceed 2 * 105.
 
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