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Diffstat (limited to 'min_operations_arr_eq.py')
-rw-r--r-- | min_operations_arr_eq.py | 35 |
1 files changed, 35 insertions, 0 deletions
diff --git a/min_operations_arr_eq.py b/min_operations_arr_eq.py new file mode 100644 index 0000000..0434365 --- /dev/null +++ b/min_operations_arr_eq.py @@ -0,0 +1,35 @@ +# 2602. Minimum Operations to Make All Array Elements Equal +from itertools import accumulate +from bisect import bisect_left + +""" +given an array 'nums' consisting of positive integers. you are also given an +integer array 'queries' of size 'm'. for the i'th query, you want to make all +of the elements of 'nums' equal to 'queries[i]'. you can perform the +following operation on the array any number of times: increase or decrease an +element of the array by 1. return an array 'ans' of size 'm' where 'ans[i]' +is the minimum number of operations to make all elements of 'nums' equal to +'queries[i]'. note that after each query the array is reset to its original +state +""" + + +class Solution(object): + def minOperations(self, nums, queries): + """ + :type nums: List[int] + :type queries: List[int] + :rtype: List[int] + """ + nums.sort() + ans, n, prefix = [], len(nums), [0] + list(accumulate(nums)) + for i in queries: + j = bisect_left(nums, i) + ans.append(i * (2 * j - n) + prefix[n] - 2 * prefix[j]) + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.minOperations(nums=[3, 1, 6, 8], queries=[1, 5])) + print(obj.minOperations(nums=[2, 9, 6, 3], queries=[10])) |