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You are given two arrays nums and andValues of length n and m respectively. The value of an array is equal to the last element of that array. You have to divide nums into m disjoint contiguous subarrays such that for the ith subarray [li, ri], the bitwise AND of the subarray elements is equal to andValues[i], in other words, nums[li] & nums[li + 1] & ... & nums[ri] == andValues[i] for all 1 <= i <= m, where & represents the bitwise AND operator. Return the minimum possible sum of the values of the m subarrays nums is divided into. If it is not possible to divide nums into m subarrays satisfying these conditions, return -1.
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You are given two arrays nums and andValues of length n and m respectively. The value of an array is equal to the last element of that array. You have to divide nums into m disjoint contiguous subarrays such that for the ith subarray [li, ri], the bitwise AND of the subarray elements is equal to andValues[i], in other words, nums[li] & nums[li + 1] & ... & nums[ri] == andValues[i] for all 1 <= i <= m, where & represents the bitwise AND operator. Return the minimum possible sum of the values of the m subarrays nums is divided into. If it is not possible to divide nums into m subarrays satisfying these conditions, return -1.
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nums = [1,4,3,3,2], andValues = [0,3,3,2]
12
nums = [2,3,5,7,7,7,5], andValues = [0,7,5]
17
nums = [1,2,3,4], andValues = [2]
-1
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Solve Minimum Sum of Values by Dividing Array — You are given two arrays nums and andValues of length n and m respectively. The ...
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Time: O(n) | Space: O(n)
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