The sum of n, n-1, n-2, and n-3 is 4n-6.
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The minimum unique array sum that can be achieved is when all elements in the array are different, resulting in the sum of the array being equal to the sum of the first n natural numbers, which is n(n1)/2.
To show that the solution of the recurrence relation t(n) t(n-1) n is in O(n2), we can use the Master Theorem. This theorem helps analyze the time complexity of recursive algorithms. In this case, the recurrence relation can be seen as T(n) T(n-1) n, which falls under the Master Theorem's first case where a 1, b 1, and f(n) n. Since f(n) n is polynomially larger than nlogb(a) n0, the solution is in O(n2).
The time complexity of the algorithm represented by the recurrence relation t(n) 4t(n/2) n2 logn is O(n2 log2 n).
The function t(n) relates to the function t(n1/2) 1 by taking the square root of n in the second function and adding 1 to the result.
The total running time of counting from 1 to n in binary is log(n1) steps.