https://hack.codingblocks.com/app/contests/16/966/problem
In this solution two for loops are used but i couldnt understand the use of two for loops I removed the first for loop and ran the code it showed me out of two tedt cases ome failed but since this test case contains 500 strings i am unable to deduce upto any conclusion please help at what point my one test case fails.
@AyushKumar9032 hey isme bhar wala loop pure string array ko traverse krne ke lie hai aur andr wala har element ko compare krega islie woh use horha hai ,so hme puri array to traverse krni pdegi islie hm use krrhe hai ye.
I<n-i-1
Is condition Mae n-1 to smj Aya pr n-i-1 nyi pta chal Raha jo apne upr Kaha please aap use ek test case se please bta dijye. I will be very much thankful to you
@ayush693 hn its like bubble sort , dekho hm upar wala loop to n-1 times chla rhe hai woh number of passes ke lie hai ,aur andr wala loop hmare elements ko compare krne ke lie ye n-i-1 islie chl rha hai kyuunki har pass me apna akhri element sahi position pr ajayega suppose for i=0 andr wala loop n-1 ytak chlega ye compare krne ke bad ab i=1 hoga fir ye n-1-1 that is n-2 tk chlega kyunki akhri element sahi position pr achuka hai islie usse ek kam tk chalyaenge andr wala loop,see eg:
this ex is on number array but same logic is for string array bs comapreto hmne apna bnaya hai:
Example:
First Pass:
( 5 1 4 2 8 ) –> ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps since 5 > 1.
( 1 5 4 2 8 ) –> ( 1 4 5 2 8 ), Swap since 5 > 4
( 1 4 5 2 8 ) –> ( 1 4 2 5 8 ), Swap since 5 > 2
( 1 4 2 5 8 ) –> ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them.
Second Pass:
( 1 4 2 5 8 ) –> ( 1 4 2 5 8 )
( 1 4 2 5 8 ) –> ( 1 2 4 5 8 ), Swap since 4 > 2
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.
Third Pass:
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )