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can you please tell me how did u come to the conclusion it is a average case ? is average case = best case ??
hi,i have a doubt in this. For j=1 to n-1[LIST] [*]2 to n [*]3 to n [*]4 to n [*]..... [*]n to n[/LIST]it should be like this right ? For j=1 to n-1[LIST] [*]2 to n [*]3 to n [*]4 to n [*]..... [*]n-1 to n[/LIST]when j=n-1 , k's value …
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