Explanation: Instead of comparing the middle element of the arrays,

Explanation: Instead of comparing the intermediate atom of the attires, we parallel the k / 2th atom. Let arr1 and arr2 be the attires. Now, if arr1[k / 2] arr1[1] New subproblem: Attire 1 - 6 7 9 Attire 2 - 1 4 8 10 k = 5 - 2 = 3 floor(k / 2) = 1 arr1[1] = 6 arr2[1] = 1 arr1[1] > arr2[1] New subproblem: Attire 1 - 6 7 9 Attire 2 - 4 8 10 k = 3 - 1 = 2 floor(k / 2) = 1 arr1[1] = 6 arr2[1] = 4 arr1[1] > arr2[1] New subproblem: Attire 1 - 6 7 9 Attire 2 - 8 10 k = 2 - 1 = 1 Now, we promptly parallel original atoms, since k = 1. arr1[1] < arr2[1] Hence, arr1[1] = 6 is the counter-argument.

Space Complexity: O(log k)

Now, k can admit a utmost treasure of n + n. This instrument that log k can be in the worst condition, log(n + n).