Vwwxrf

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Smoothness of finite-dimensional functional calculus

How can bays and straits be determined in a procedurally generated map?

Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?

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Adding span tags within wp_list_pages list items

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What do the dots in this tr command do: tr .............A-Z A-ZA-Z <<< "JVPQBOV" (with 13 dots)

Mathematical cryptic clues

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Minimum swaps algorithm terminated due to timeout

SPOJ GENERAL: sorting by swaps of distance kHackerrank Insertion Sort Algorithm 1 (creating duplicates to show shifting)Terminated due to timeout Mapping phone numbersDetermining the cardinality of sets defined by recursively following indexesSearching the Array Index & Element EqualityK Messed Array sort in JavaClimbing the Leaderboard: HackerranK, Terminated due to timeoutMinimum number of swaps required to sort the array in ascending orderTwo-sum solution in JavaScriptFind array index with element equality in Python

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

4

$begingroup$

I have been trying to solve this question.

Given an unordered array consisting of consecutive integers [1, 2, 3, …, n], find the minimum number of two-element swaps to sort the array.

I was able to solve it but my solution's complexity is not good enough that it terminated due to a timeout when it runs with bigger arrays. This is a typical problem for me, I always solve the problem somehow but the complexity is not optimal and the solution does not pass the all the cast cases. If you have any suggestions for me for the future interview questions like that, I'd appreciate knowing how should I approach the questions.

function findIndice(arr,i)
 let iterator=0
 while(arr[iterator]!==i+1)
 iterator++
 
 return iterator


function swap(arr,x,y)
 let temp=arr[x]
 arr[x]=arr[y]
 arr[y]=temp




function minimumSwaps(arr)
 let i=0;
 let counter=0;
 let size=arr.length;

 for(i=0;i<size-1;i++)
 if(arr[i]!==i+1)
 let index=findIndice(arr,i)
 swap(arr,index,i)
 counter++
 
 
 return counter

javascript algorithm sorting time-limit-exceeded interview-questions

share|improve this question

edited Aug 11 '18 at 22:12

200_success

131k17157422

asked Aug 11 '18 at 17:11

Ugur YilmazUgur Yilmaz

413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a totally different question... @Gerrit0
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 2:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh shoot, you are totally right. I must have sent the wrong link as I looked at several... sorry about that!
  $endgroup$
  – Gerrit0
  Aug 12 '18 at 2:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's okay, Thanks for trying to help :) @Gerrit0
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 3:50
  

  
  
  
  

add a comment | 

4

$begingroup$

I have been trying to solve this question.

Given an unordered array consisting of consecutive integers [1, 2, 3, …, n], find the minimum number of two-element swaps to sort the array.

I was able to solve it but my solution's complexity is not good enough that it terminated due to a timeout when it runs with bigger arrays. This is a typical problem for me, I always solve the problem somehow but the complexity is not optimal and the solution does not pass the all the cast cases. If you have any suggestions for me for the future interview questions like that, I'd appreciate knowing how should I approach the questions.

function findIndice(arr,i)
 let iterator=0
 while(arr[iterator]!==i+1)
 iterator++
 
 return iterator


function swap(arr,x,y)
 let temp=arr[x]
 arr[x]=arr[y]
 arr[y]=temp




function minimumSwaps(arr)
 let i=0;
 let counter=0;
 let size=arr.length;

 for(i=0;i<size-1;i++)
 if(arr[i]!==i+1)
 let index=findIndice(arr,i)
 swap(arr,index,i)
 counter++
 
 
 return counter

javascript algorithm sorting time-limit-exceeded interview-questions

share|improve this question

edited Aug 11 '18 at 22:12

200_success

131k17157422

asked Aug 11 '18 at 17:11

Ugur YilmazUgur Yilmaz

413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a totally different question... @Gerrit0
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 2:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh shoot, you are totally right. I must have sent the wrong link as I looked at several... sorry about that!
  $endgroup$
  – Gerrit0
  Aug 12 '18 at 2:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's okay, Thanks for trying to help :) @Gerrit0
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 3:50
  

  
  
  
  

add a comment | 

$begingroup$

I have been trying to solve this question.

Given an unordered array consisting of consecutive integers [1, 2, 3, …, n], find the minimum number of two-element swaps to sort the array.

I was able to solve it but my solution's complexity is not good enough that it terminated due to a timeout when it runs with bigger arrays. This is a typical problem for me, I always solve the problem somehow but the complexity is not optimal and the solution does not pass the all the cast cases. If you have any suggestions for me for the future interview questions like that, I'd appreciate knowing how should I approach the questions.

function findIndice(arr,i)
 let iterator=0
 while(arr[iterator]!==i+1)
 iterator++
 
 return iterator


function swap(arr,x,y)
 let temp=arr[x]
 arr[x]=arr[y]
 arr[y]=temp




function minimumSwaps(arr)
 let i=0;
 let counter=0;
 let size=arr.length;

 for(i=0;i<size-1;i++)
 if(arr[i]!==i+1)
 let index=findIndice(arr,i)
 swap(arr,index,i)
 counter++
 
 
 return counter

javascript algorithm sorting time-limit-exceeded interview-questions

share|improve this question

edited Aug 11 '18 at 22:12

200_success

131k17157422

asked Aug 11 '18 at 17:11

Ugur YilmazUgur Yilmaz

413

$endgroup$

function findIndice(arr,i)
 let iterator=0
 while(arr[iterator]!==i+1)
 iterator++
 
 return iterator


function swap(arr,x,y)
 let temp=arr[x]
 arr[x]=arr[y]
 arr[y]=temp




function minimumSwaps(arr)
 let i=0;
 let counter=0;
 let size=arr.length;

 for(i=0;i<size-1;i++)
 if(arr[i]!==i+1)
 let index=findIndice(arr,i)
 swap(arr,index,i)
 counter++
 
 
 return counter

share|improve this question

edited Aug 11 '18 at 22:12

200_success

131k17157422

asked Aug 11 '18 at 17:11

Ugur YilmazUgur Yilmaz

413

share|improve this question

edited Aug 11 '18 at 22:12

200_success

131k17157422

asked Aug 11 '18 at 17:11

Ugur YilmazUgur Yilmaz

413

edited Aug 11 '18 at 22:12

200_success

131k17157422

edited Aug 11 '18 at 22:12

200_success

131k17157422

asked Aug 11 '18 at 17:11

Ugur YilmazUgur Yilmaz

413

asked Aug 11 '18 at 17:11

Ugur YilmazUgur Yilmaz

413

2 Answers
2

active

oldest

votes

3

$begingroup$

Your algorithm can be summarized by the pseudocode:

for each position in the array
 if a position is occupied by the wrong number
 find the number that fits into the position
 perform a swap

A better algorithm would be:

for each position in the array
 if a number is in the wrong location
 find the position the number should go
 perform a swap

This is because finding a number for a given location requires a linear scan, but finding the location for a given number is as simple as it gets: the number five should go into the fifth position.

The full program could look like this (in python, as I don't speak js):

def minimumSwaps(arr):
 def swap(i, j):
 arr[i], arr[j] = arr[j], arr[i]
 swaps = 0
 for i in range(0, len(arr)):
 while arr[i]-1 != i:
 swap(i, arr[i]-1)
 swaps += 1
 return swaps

share|improve this answer

answered Aug 12 '18 at 10:25

Rainer P.Rainer P.

1,261414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a great solution but it fails if the numbers are not consecutive. To illustrate; 1 3 5 2 4 6 8 does not pass this test case @Rainer P.
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @UgurYilmaz - Your question literaly says [...] consisting of consecutive integers [...].
  $endgroup$
  – Rainer P.
  Aug 12 '18 at 18:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know but there should be an error in hackerRank, then. Because there is a test case with that input and it seems to fail. Do you have any idea how to implement a solution efficiently if it is not a consecutive array integers @Rainer P.
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 19:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There aren't such cases, I tried the solution above and it was accepted.
  $endgroup$
  – user158037
  Oct 12 '18 at 16:07
  

  
  
  
  

add a comment | 

0

$begingroup$

The fastest way so far I tried is

1. For each iteration of "i"


2. Try to find out from the list ( i - 1, arr.length ) if there is an item misplaced


3. If there is an misplaced item, swap it

const minimalSwap = (arr) => 
 // don't mutate the original array
 let list = [...arr]
 // length of the list
 const listLen = list.length
 // state - total swap
 let totalSwap = 0
 if (listLen <= 1) 
 return totalSwap
 
 function swap(items, idxOne, idxTwo) 
 let tempNode = items[idxOne]
 items[idxOne] = items[idxTwo]
 items[idxTwo] = tempNode
 totalSwap += 1
 return items
 
 for (let i = 0; i < listLen; i++) 
 // for each iteration, we find the
 // number that should go to the correct position
 let idToSwap = false
 const correctNumber = i + 1
 // iterate through starting next item
 // and find if the correct number is misplaced
 for (let j = i + 1; j < listLen; j++) 
 // if it is misplaced, swap it
 if (correctNumber === list[j]) 
 idToSwap = j
 break;
 
 
 if (idToSwap) 
 list = swap(list, i, idToSwap)
 
 

 return totalSwap


// test case
minimalSwap([14, 15, 16, 4, 8, 3, 1, 2, 5, 7, 9, 10, 11, 12, 13, 6])

share

answered 20 secs ago

R.RR.R

101

New contributor

R.R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

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3

$begingroup$

Your algorithm can be summarized by the pseudocode:

for each position in the array
 if a position is occupied by the wrong number
 find the number that fits into the position
 perform a swap

A better algorithm would be:

for each position in the array
 if a number is in the wrong location
 find the position the number should go
 perform a swap

This is because finding a number for a given location requires a linear scan, but finding the location for a given number is as simple as it gets: the number five should go into the fifth position.

The full program could look like this (in python, as I don't speak js):

def minimumSwaps(arr):
 def swap(i, j):
 arr[i], arr[j] = arr[j], arr[i]
 swaps = 0
 for i in range(0, len(arr)):
 while arr[i]-1 != i:
 swap(i, arr[i]-1)
 swaps += 1
 return swaps

share|improve this answer

answered Aug 12 '18 at 10:25

Rainer P.Rainer P.

1,261414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a great solution but it fails if the numbers are not consecutive. To illustrate; 1 3 5 2 4 6 8 does not pass this test case @Rainer P.
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @UgurYilmaz - Your question literaly says [...] consisting of consecutive integers [...].
  $endgroup$
  – Rainer P.
  Aug 12 '18 at 18:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know but there should be an error in hackerRank, then. Because there is a test case with that input and it seems to fail. Do you have any idea how to implement a solution efficiently if it is not a consecutive array integers @Rainer P.
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 19:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There aren't such cases, I tried the solution above and it was accepted.
  $endgroup$
  – user158037
  Oct 12 '18 at 16:07
  

  
  
  
  

add a comment | 

3

$begingroup$

Your algorithm can be summarized by the pseudocode:

for each position in the array
 if a position is occupied by the wrong number
 find the number that fits into the position
 perform a swap

A better algorithm would be:

for each position in the array
 if a number is in the wrong location
 find the position the number should go
 perform a swap

This is because finding a number for a given location requires a linear scan, but finding the location for a given number is as simple as it gets: the number five should go into the fifth position.

The full program could look like this (in python, as I don't speak js):

def minimumSwaps(arr):
 def swap(i, j):
 arr[i], arr[j] = arr[j], arr[i]
 swaps = 0
 for i in range(0, len(arr)):
 while arr[i]-1 != i:
 swap(i, arr[i]-1)
 swaps += 1
 return swaps

share|improve this answer

answered Aug 12 '18 at 10:25

Rainer P.Rainer P.

1,261414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a great solution but it fails if the numbers are not consecutive. To illustrate; 1 3 5 2 4 6 8 does not pass this test case @Rainer P.
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @UgurYilmaz - Your question literaly says [...] consisting of consecutive integers [...].
  $endgroup$
  – Rainer P.
  Aug 12 '18 at 18:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know but there should be an error in hackerRank, then. Because there is a test case with that input and it seems to fail. Do you have any idea how to implement a solution efficiently if it is not a consecutive array integers @Rainer P.
  $endgroup$
  – Ugur Yilmaz
  Aug 12 '18 at 19:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There aren't such cases, I tried the solution above and it was accepted.
  $endgroup$
  – user158037
  Oct 12 '18 at 16:07
  

  
  
  
  

add a comment | 

$begingroup$

Your algorithm can be summarized by the pseudocode:

for each position in the array
 if a position is occupied by the wrong number
 find the number that fits into the position
 perform a swap

A better algorithm would be:

for each position in the array
 if a number is in the wrong location
 find the position the number should go
 perform a swap

This is because finding a number for a given location requires a linear scan, but finding the location for a given number is as simple as it gets: the number five should go into the fifth position.

The full program could look like this (in python, as I don't speak js):

def minimumSwaps(arr):
 def swap(i, j):
 arr[i], arr[j] = arr[j], arr[i]
 swaps = 0
 for i in range(0, len(arr)):
 while arr[i]-1 != i:
 swap(i, arr[i]-1)
 swaps += 1
 return swaps

share|improve this answer

answered Aug 12 '18 at 10:25

Rainer P.Rainer P.

1,261414

$endgroup$

for each position in the array
 if a position is occupied by the wrong number
 find the number that fits into the position
 perform a swap

for each position in the array
 if a number is in the wrong location
 find the position the number should go
 perform a swap

def minimumSwaps(arr):
 def swap(i, j):
 arr[i], arr[j] = arr[j], arr[i]
 swaps = 0
 for i in range(0, len(arr)):
 while arr[i]-1 != i:
 swap(i, arr[i]-1)
 swaps += 1
 return swaps

share|improve this answer

answered Aug 12 '18 at 10:25

Rainer P.Rainer P.

1,261414

answered Aug 12 '18 at 10:25

Rainer P.Rainer P.

1,261414

answered Aug 12 '18 at 10:25

Rainer P.Rainer P.

1,261414

0

$begingroup$

The fastest way so far I tried is

1. For each iteration of "i"


2. Try to find out from the list ( i - 1, arr.length ) if there is an item misplaced


3. If there is an misplaced item, swap it

const minimalSwap = (arr) => 
 // don't mutate the original array
 let list = [...arr]
 // length of the list
 const listLen = list.length
 // state - total swap
 let totalSwap = 0
 if (listLen <= 1) 
 return totalSwap
 
 function swap(items, idxOne, idxTwo) 
 let tempNode = items[idxOne]
 items[idxOne] = items[idxTwo]
 items[idxTwo] = tempNode
 totalSwap += 1
 return items
 
 for (let i = 0; i < listLen; i++) 
 // for each iteration, we find the
 // number that should go to the correct position
 let idToSwap = false
 const correctNumber = i + 1
 // iterate through starting next item
 // and find if the correct number is misplaced
 for (let j = i + 1; j < listLen; j++) 
 // if it is misplaced, swap it
 if (correctNumber === list[j]) 
 idToSwap = j
 break;
 
 
 if (idToSwap) 
 list = swap(list, i, idToSwap)
 
 

 return totalSwap


// test case
minimalSwap([14, 15, 16, 4, 8, 3, 1, 2, 5, 7, 9, 10, 11, 12, 13, 6])

share

answered 20 secs ago

R.RR.R

101

New contributor

R.R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

The fastest way so far I tried is

1. For each iteration of "i"


2. Try to find out from the list ( i - 1, arr.length ) if there is an item misplaced


3. If there is an misplaced item, swap it

const minimalSwap = (arr) => 
 // don't mutate the original array
 let list = [...arr]
 // length of the list
 const listLen = list.length
 // state - total swap
 let totalSwap = 0
 if (listLen <= 1) 
 return totalSwap
 
 function swap(items, idxOne, idxTwo) 
 let tempNode = items[idxOne]
 items[idxOne] = items[idxTwo]
 items[idxTwo] = tempNode
 totalSwap += 1
 return items
 
 for (let i = 0; i < listLen; i++) 
 // for each iteration, we find the
 // number that should go to the correct position
 let idToSwap = false
 const correctNumber = i + 1
 // iterate through starting next item
 // and find if the correct number is misplaced
 for (let j = i + 1; j < listLen; j++) 
 // if it is misplaced, swap it
 if (correctNumber === list[j]) 
 idToSwap = j
 break;
 
 
 if (idToSwap) 
 list = swap(list, i, idToSwap)
 
 

 return totalSwap


// test case
minimalSwap([14, 15, 16, 4, 8, 3, 1, 2, 5, 7, 9, 10, 11, 12, 13, 6])

share

answered 20 secs ago

R.RR.R

101

New contributor

R.R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

The fastest way so far I tried is

1. For each iteration of "i"


2. Try to find out from the list ( i - 1, arr.length ) if there is an item misplaced


3. If there is an misplaced item, swap it

const minimalSwap = (arr) => 
 // don't mutate the original array
 let list = [...arr]
 // length of the list
 const listLen = list.length
 // state - total swap
 let totalSwap = 0
 if (listLen <= 1) 
 return totalSwap
 
 function swap(items, idxOne, idxTwo) 
 let tempNode = items[idxOne]
 items[idxOne] = items[idxTwo]
 items[idxTwo] = tempNode
 totalSwap += 1
 return items
 
 for (let i = 0; i < listLen; i++) 
 // for each iteration, we find the
 // number that should go to the correct position
 let idToSwap = false
 const correctNumber = i + 1
 // iterate through starting next item
 // and find if the correct number is misplaced
 for (let j = i + 1; j < listLen; j++) 
 // if it is misplaced, swap it
 if (correctNumber === list[j]) 
 idToSwap = j
 break;
 
 
 if (idToSwap) 
 list = swap(list, i, idToSwap)
 
 

 return totalSwap


// test case
minimalSwap([14, 15, 16, 4, 8, 3, 1, 2, 5, 7, 9, 10, 11, 12, 13, 6])

share

answered 20 secs ago

R.RR.R

101

New contributor

R.R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

const minimalSwap = (arr) => 
 // don't mutate the original array
 let list = [...arr]
 // length of the list
 const listLen = list.length
 // state - total swap
 let totalSwap = 0
 if (listLen <= 1) 
 return totalSwap
 
 function swap(items, idxOne, idxTwo) 
 let tempNode = items[idxOne]
 items[idxOne] = items[idxTwo]
 items[idxTwo] = tempNode
 totalSwap += 1
 return items
 
 for (let i = 0; i < listLen; i++) 
 // for each iteration, we find the
 // number that should go to the correct position
 let idToSwap = false
 const correctNumber = i + 1
 // iterate through starting next item
 // and find if the correct number is misplaced
 for (let j = i + 1; j < listLen; j++) 
 // if it is misplaced, swap it
 if (correctNumber === list[j]) 
 idToSwap = j
 break;
 
 
 if (idToSwap) 
 list = swap(list, i, idToSwap)
 
 

 return totalSwap


// test case
minimalSwap([14, 15, 16, 4, 8, 3, 1, 2, 5, 7, 9, 10, 11, 12, 13, 6])

const minimalSwap = (arr) => 
 // don't mutate the original array
 let list = [...arr]
 // length of the list
 const listLen = list.length
 // state - total swap
 let totalSwap = 0
 if (listLen <= 1) 
 return totalSwap
 
 function swap(items, idxOne, idxTwo) 
 let tempNode = items[idxOne]
 items[idxOne] = items[idxTwo]
 items[idxTwo] = tempNode
 totalSwap += 1
 return items
 
 for (let i = 0; i < listLen; i++) 
 // for each iteration, we find the
 // number that should go to the correct position
 let idToSwap = false
 const correctNumber = i + 1
 // iterate through starting next item
 // and find if the correct number is misplaced
 for (let j = i + 1; j < listLen; j++) 
 // if it is misplaced, swap it
 if (correctNumber === list[j]) 
 idToSwap = j
 break;
 
 
 if (idToSwap) 
 list = swap(list, i, idToSwap)
 
 

 return totalSwap


// test case
minimalSwap([14, 15, 16, 4, 8, 3, 1, 2, 5, 7, 9, 10, 11, 12, 13, 6])

const minimalSwap = (arr) => 
 // don't mutate the original array
 let list = [...arr]
 // length of the list
 const listLen = list.length
 // state - total swap
 let totalSwap = 0
 if (listLen <= 1) 
 return totalSwap
 
 function swap(items, idxOne, idxTwo) 
 let tempNode = items[idxOne]
 items[idxOne] = items[idxTwo]
 items[idxTwo] = tempNode
 totalSwap += 1
 return items
 
 for (let i = 0; i < listLen; i++) 
 // for each iteration, we find the
 // number that should go to the correct position
 let idToSwap = false
 const correctNumber = i + 1
 // iterate through starting next item
 // and find if the correct number is misplaced
 for (let j = i + 1; j < listLen; j++) 
 // if it is misplaced, swap it
 if (correctNumber === list[j]) 
 idToSwap = j
 break;
 
 
 if (idToSwap) 
 list = swap(list, i, idToSwap)
 
 

 return totalSwap


// test case
minimalSwap([14, 15, 16, 4, 8, 3, 1, 2, 5, 7, 9, 10, 11, 12, 13, 6])

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answered 20 secs ago

R.RR.R

101

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answered 20 secs ago

R.RR.R

101

New contributor

R.R is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 20 secs ago

R.RR.R

101