Searching in sorted and rotated

#1

for mid lies in the left line
if(a[s]<=a[mid){
}
this case has to be satisfied
i am not able to understand it

#2

Hello @Chetan_Khandelwal,

Please share the link of this video or content.
I cannot access it.

#3
#4

Hello @Chetan_Khandelwal,

As sir explained in the video:

  1. Initially you had a sorted array in increasing order.
    Suppose: 0 1 2 3 4 5 6 7 8

  2. Now you have rotated the array towards right by 3 elements:
    6 7 8 0 1 2 3 4 5

  3. Now, observe the resulted sequence,
    there are two increasing functions:
    first, 6 7 8
    second, 01 2 3 4 5

  4. So, after calculating mid=(low+high)/2 =(0+8)/2=4 i.e. third index: a[mid]=1.
    Now, you have to check which sequence among-st two specified in point 3, mid lies in.
    4.1. if a[low]<a[mid]: then mid lies on the first sequence
    this is because all the elements bigger than a[low] lies in left sequence
    4.2. ielsef a[mid]<a[high]: then mid lies on the second sequence.
    because all the elements smaller than a[high] lies in second/left sequence

Hope, this would help.
Give a like, if you are satisfied.

1 Like
#5

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