Count pairs in list of integers such that their addition is equal to the input value
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Given a List<int>
, the problem I am trying to solve is:
Find the number of unique pairs in List<int>
such that their addition is exactly equal to the input value.
The bottleneck is a nested for()
loop that I have used to go through the list
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) &&
IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
How can this be made better for performance? I understand 'better' is a subjective word and has no real meaning as such. For a List<int>
of size 50000 the code takes about 18 seconds on one of my VM. For 500000 it's way too worst. So, here by 'better' I mean 'faster'. Clearly this problem deserves much less than 18 seconds to solve in my opinion. With 1 Parallel.For()
I have managed to get the loop time to 10-11 seconds, but I have a feeling that this whole algorithm needs a fresh set of eyes to look at.
Parallel.For(0, intList.Count - 1,
i =>
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
);
How can I speed this up?
Full code from my console application is as below:
class TestClass
static List<int> compareList = new List<int>();
// GetPossibleCombination method.
// This method finds out the unique number of possible combinations where
// addition of any 2 values from the list is exactly equal to 'totalValue'
static int GetPossibleCombinations(List<int> intList, long totalValue)
totalValue > 5000000000)
return 0;
compareList.Clear();
int nCombinations = 0;
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++) // start from this element onwards
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
return nCombinations;
// This method creates a list of possible values we have
static bool IsPairUnique(int v1, int v2)
static void Main(string args)
int intListSize = 50000; // Optimize it for numbers upto 500,000
long totalValue = 5000;
List<int> intList = new List<int>();
Random r = new Random();
for (int i = 0; i < intListSize; i++)
intList.Add(r.Next(0, 10000)); // populate random values.
Stopwatch sw = new Stopwatch();
sw.Start();
// Find the number of unique pairs in 'intList' such that
// their addition is exactly equal to 'totalValue'
int res = GetPossibleCombinations(intList, totalValue);
sw.Stop();
Console.WriteLine(sw.Elapsed.ToString());
c# performance programming-challenge k-sum
$endgroup$
add a comment |
$begingroup$
Given a List<int>
, the problem I am trying to solve is:
Find the number of unique pairs in List<int>
such that their addition is exactly equal to the input value.
The bottleneck is a nested for()
loop that I have used to go through the list
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) &&
IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
How can this be made better for performance? I understand 'better' is a subjective word and has no real meaning as such. For a List<int>
of size 50000 the code takes about 18 seconds on one of my VM. For 500000 it's way too worst. So, here by 'better' I mean 'faster'. Clearly this problem deserves much less than 18 seconds to solve in my opinion. With 1 Parallel.For()
I have managed to get the loop time to 10-11 seconds, but I have a feeling that this whole algorithm needs a fresh set of eyes to look at.
Parallel.For(0, intList.Count - 1,
i =>
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
);
How can I speed this up?
Full code from my console application is as below:
class TestClass
static List<int> compareList = new List<int>();
// GetPossibleCombination method.
// This method finds out the unique number of possible combinations where
// addition of any 2 values from the list is exactly equal to 'totalValue'
static int GetPossibleCombinations(List<int> intList, long totalValue)
totalValue > 5000000000)
return 0;
compareList.Clear();
int nCombinations = 0;
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++) // start from this element onwards
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
return nCombinations;
// This method creates a list of possible values we have
static bool IsPairUnique(int v1, int v2)
static void Main(string args)
int intListSize = 50000; // Optimize it for numbers upto 500,000
long totalValue = 5000;
List<int> intList = new List<int>();
Random r = new Random();
for (int i = 0; i < intListSize; i++)
intList.Add(r.Next(0, 10000)); // populate random values.
Stopwatch sw = new Stopwatch();
sw.Start();
// Find the number of unique pairs in 'intList' such that
// their addition is exactly equal to 'totalValue'
int res = GetPossibleCombinations(intList, totalValue);
sw.Stop();
Console.WriteLine(sw.Elapsed.ToString());
c# performance programming-challenge k-sum
$endgroup$
1
$begingroup$
Does "unique pairs" mean "unique indexes", "unique values", or "unique combinations of values"? For example: if the data is1, 1, 1
and the input value is2
, do we have zero, one or three unique pairs?
$endgroup$
– Oh My Goodness
Feb 4 at 1:13
1
$begingroup$
@OhMyGoodness, "unique pairs" mean "unique combinations of values" from the list. e.g.: If 'totalValue' we are looking for is 5 and the value at list[i] is 2, list[i+1] is 3 and list[j] 3, list[j+1] is 2 then only one pair (either i, i+1 or j, j+1) should be considered. Hope this helps.
$endgroup$
– silverspoon
Feb 4 at 1:55
add a comment |
$begingroup$
Given a List<int>
, the problem I am trying to solve is:
Find the number of unique pairs in List<int>
such that their addition is exactly equal to the input value.
The bottleneck is a nested for()
loop that I have used to go through the list
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) &&
IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
How can this be made better for performance? I understand 'better' is a subjective word and has no real meaning as such. For a List<int>
of size 50000 the code takes about 18 seconds on one of my VM. For 500000 it's way too worst. So, here by 'better' I mean 'faster'. Clearly this problem deserves much less than 18 seconds to solve in my opinion. With 1 Parallel.For()
I have managed to get the loop time to 10-11 seconds, but I have a feeling that this whole algorithm needs a fresh set of eyes to look at.
Parallel.For(0, intList.Count - 1,
i =>
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
);
How can I speed this up?
Full code from my console application is as below:
class TestClass
static List<int> compareList = new List<int>();
// GetPossibleCombination method.
// This method finds out the unique number of possible combinations where
// addition of any 2 values from the list is exactly equal to 'totalValue'
static int GetPossibleCombinations(List<int> intList, long totalValue)
totalValue > 5000000000)
return 0;
compareList.Clear();
int nCombinations = 0;
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++) // start from this element onwards
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
return nCombinations;
// This method creates a list of possible values we have
static bool IsPairUnique(int v1, int v2)
static void Main(string args)
int intListSize = 50000; // Optimize it for numbers upto 500,000
long totalValue = 5000;
List<int> intList = new List<int>();
Random r = new Random();
for (int i = 0; i < intListSize; i++)
intList.Add(r.Next(0, 10000)); // populate random values.
Stopwatch sw = new Stopwatch();
sw.Start();
// Find the number of unique pairs in 'intList' such that
// their addition is exactly equal to 'totalValue'
int res = GetPossibleCombinations(intList, totalValue);
sw.Stop();
Console.WriteLine(sw.Elapsed.ToString());
c# performance programming-challenge k-sum
$endgroup$
Given a List<int>
, the problem I am trying to solve is:
Find the number of unique pairs in List<int>
such that their addition is exactly equal to the input value.
The bottleneck is a nested for()
loop that I have used to go through the list
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) &&
IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
How can this be made better for performance? I understand 'better' is a subjective word and has no real meaning as such. For a List<int>
of size 50000 the code takes about 18 seconds on one of my VM. For 500000 it's way too worst. So, here by 'better' I mean 'faster'. Clearly this problem deserves much less than 18 seconds to solve in my opinion. With 1 Parallel.For()
I have managed to get the loop time to 10-11 seconds, but I have a feeling that this whole algorithm needs a fresh set of eyes to look at.
Parallel.For(0, intList.Count - 1,
i =>
for (int j = i + 1; j < intList.Count; j++)
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
);
How can I speed this up?
Full code from my console application is as below:
class TestClass
static List<int> compareList = new List<int>();
// GetPossibleCombination method.
// This method finds out the unique number of possible combinations where
// addition of any 2 values from the list is exactly equal to 'totalValue'
static int GetPossibleCombinations(List<int> intList, long totalValue)
totalValue > 5000000000)
return 0;
compareList.Clear();
int nCombinations = 0;
// *** OPTIMIZE THIS LOOP ***
for (int i = 0; i < intList.Count - 1; i++)
for (int j = i + 1; j < intList.Count; j++) // start from this element onwards
if ((intList[i] + intList[j] == totalValue) && IsPairUnique(intList[i], intList[j]) == true)
nCombinations++;
return nCombinations;
// This method creates a list of possible values we have
static bool IsPairUnique(int v1, int v2)
static void Main(string args)
int intListSize = 50000; // Optimize it for numbers upto 500,000
long totalValue = 5000;
List<int> intList = new List<int>();
Random r = new Random();
for (int i = 0; i < intListSize; i++)
intList.Add(r.Next(0, 10000)); // populate random values.
Stopwatch sw = new Stopwatch();
sw.Start();
// Find the number of unique pairs in 'intList' such that
// their addition is exactly equal to 'totalValue'
int res = GetPossibleCombinations(intList, totalValue);
sw.Stop();
Console.WriteLine(sw.Elapsed.ToString());
c# performance programming-challenge k-sum
c# performance programming-challenge k-sum
edited Feb 4 at 15:32
t3chb0t
34.7k750121
34.7k750121
asked Feb 4 at 1:06
silverspoonsilverspoon
1261
1261
1
$begingroup$
Does "unique pairs" mean "unique indexes", "unique values", or "unique combinations of values"? For example: if the data is1, 1, 1
and the input value is2
, do we have zero, one or three unique pairs?
$endgroup$
– Oh My Goodness
Feb 4 at 1:13
1
$begingroup$
@OhMyGoodness, "unique pairs" mean "unique combinations of values" from the list. e.g.: If 'totalValue' we are looking for is 5 and the value at list[i] is 2, list[i+1] is 3 and list[j] 3, list[j+1] is 2 then only one pair (either i, i+1 or j, j+1) should be considered. Hope this helps.
$endgroup$
– silverspoon
Feb 4 at 1:55
add a comment |
1
$begingroup$
Does "unique pairs" mean "unique indexes", "unique values", or "unique combinations of values"? For example: if the data is1, 1, 1
and the input value is2
, do we have zero, one or three unique pairs?
$endgroup$
– Oh My Goodness
Feb 4 at 1:13
1
$begingroup$
@OhMyGoodness, "unique pairs" mean "unique combinations of values" from the list. e.g.: If 'totalValue' we are looking for is 5 and the value at list[i] is 2, list[i+1] is 3 and list[j] 3, list[j+1] is 2 then only one pair (either i, i+1 or j, j+1) should be considered. Hope this helps.
$endgroup$
– silverspoon
Feb 4 at 1:55
1
1
$begingroup$
Does "unique pairs" mean "unique indexes", "unique values", or "unique combinations of values"? For example: if the data is
1, 1, 1
and the input value is 2
, do we have zero, one or three unique pairs?$endgroup$
– Oh My Goodness
Feb 4 at 1:13
$begingroup$
Does "unique pairs" mean "unique indexes", "unique values", or "unique combinations of values"? For example: if the data is
1, 1, 1
and the input value is 2
, do we have zero, one or three unique pairs?$endgroup$
– Oh My Goodness
Feb 4 at 1:13
1
1
$begingroup$
@OhMyGoodness, "unique pairs" mean "unique combinations of values" from the list. e.g.: If 'totalValue' we are looking for is 5 and the value at list[i] is 2, list[i+1] is 3 and list[j] 3, list[j+1] is 2 then only one pair (either i, i+1 or j, j+1) should be considered. Hope this helps.
$endgroup$
– silverspoon
Feb 4 at 1:55
$begingroup$
@OhMyGoodness, "unique pairs" mean "unique combinations of values" from the list. e.g.: If 'totalValue' we are looking for is 5 and the value at list[i] is 2, list[i+1] is 3 and list[j] 3, list[j+1] is 2 then only one pair (either i, i+1 or j, j+1) should be considered. Hope this helps.
$endgroup$
– silverspoon
Feb 4 at 1:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The solution has a quadratic time complexity. If it solves the 50000-strong list in 18 seconds, I'd expect about 1800 seconds (aka 30 minutes) for a 500000 one. Notice that the problem is aggravated by the way you are looking for duplicates (and IsPairUnique
doesn't look correct; it is prone to false positives).
There are few standard ways to bring the complexity down.
Sort the list. Arrange two iterators, one at the beginning, another at the end.
Now add the pointed elements. If the sum is less than the target, advance the left one. If it is greater, move the right one toward the beginning. If it is a target, record it, and move both. In any case, when moving an iterator, skip duplicates. Rinse and repeat until they met.
If you have enough extra memory, put them in a set. For each element in a set check if its complement is also there.
$endgroup$
$begingroup$
The set approach won't detect 1+1=2
$endgroup$
– Oh My Goodness
Feb 4 at 2:45
$begingroup$
You could very well keep track of how many entries with a value oftotalValue/2
you have (iftotalValue
is even) - if you have 2+ entries, count it as an additional pair. For all other numbers, it doesn't matter how many of them you've seen.
$endgroup$
– mabako
Feb 4 at 12:51
add a comment |
Your Answer
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
The solution has a quadratic time complexity. If it solves the 50000-strong list in 18 seconds, I'd expect about 1800 seconds (aka 30 minutes) for a 500000 one. Notice that the problem is aggravated by the way you are looking for duplicates (and IsPairUnique
doesn't look correct; it is prone to false positives).
There are few standard ways to bring the complexity down.
Sort the list. Arrange two iterators, one at the beginning, another at the end.
Now add the pointed elements. If the sum is less than the target, advance the left one. If it is greater, move the right one toward the beginning. If it is a target, record it, and move both. In any case, when moving an iterator, skip duplicates. Rinse and repeat until they met.
If you have enough extra memory, put them in a set. For each element in a set check if its complement is also there.
$endgroup$
$begingroup$
The set approach won't detect 1+1=2
$endgroup$
– Oh My Goodness
Feb 4 at 2:45
$begingroup$
You could very well keep track of how many entries with a value oftotalValue/2
you have (iftotalValue
is even) - if you have 2+ entries, count it as an additional pair. For all other numbers, it doesn't matter how many of them you've seen.
$endgroup$
– mabako
Feb 4 at 12:51
add a comment |
$begingroup$
The solution has a quadratic time complexity. If it solves the 50000-strong list in 18 seconds, I'd expect about 1800 seconds (aka 30 minutes) for a 500000 one. Notice that the problem is aggravated by the way you are looking for duplicates (and IsPairUnique
doesn't look correct; it is prone to false positives).
There are few standard ways to bring the complexity down.
Sort the list. Arrange two iterators, one at the beginning, another at the end.
Now add the pointed elements. If the sum is less than the target, advance the left one. If it is greater, move the right one toward the beginning. If it is a target, record it, and move both. In any case, when moving an iterator, skip duplicates. Rinse and repeat until they met.
If you have enough extra memory, put them in a set. For each element in a set check if its complement is also there.
$endgroup$
$begingroup$
The set approach won't detect 1+1=2
$endgroup$
– Oh My Goodness
Feb 4 at 2:45
$begingroup$
You could very well keep track of how many entries with a value oftotalValue/2
you have (iftotalValue
is even) - if you have 2+ entries, count it as an additional pair. For all other numbers, it doesn't matter how many of them you've seen.
$endgroup$
– mabako
Feb 4 at 12:51
add a comment |
$begingroup$
The solution has a quadratic time complexity. If it solves the 50000-strong list in 18 seconds, I'd expect about 1800 seconds (aka 30 minutes) for a 500000 one. Notice that the problem is aggravated by the way you are looking for duplicates (and IsPairUnique
doesn't look correct; it is prone to false positives).
There are few standard ways to bring the complexity down.
Sort the list. Arrange two iterators, one at the beginning, another at the end.
Now add the pointed elements. If the sum is less than the target, advance the left one. If it is greater, move the right one toward the beginning. If it is a target, record it, and move both. In any case, when moving an iterator, skip duplicates. Rinse and repeat until they met.
If you have enough extra memory, put them in a set. For each element in a set check if its complement is also there.
$endgroup$
The solution has a quadratic time complexity. If it solves the 50000-strong list in 18 seconds, I'd expect about 1800 seconds (aka 30 minutes) for a 500000 one. Notice that the problem is aggravated by the way you are looking for duplicates (and IsPairUnique
doesn't look correct; it is prone to false positives).
There are few standard ways to bring the complexity down.
Sort the list. Arrange two iterators, one at the beginning, another at the end.
Now add the pointed elements. If the sum is less than the target, advance the left one. If it is greater, move the right one toward the beginning. If it is a target, record it, and move both. In any case, when moving an iterator, skip duplicates. Rinse and repeat until they met.
If you have enough extra memory, put them in a set. For each element in a set check if its complement is also there.
answered Feb 4 at 2:31
vnpvnp
40k232102
40k232102
$begingroup$
The set approach won't detect 1+1=2
$endgroup$
– Oh My Goodness
Feb 4 at 2:45
$begingroup$
You could very well keep track of how many entries with a value oftotalValue/2
you have (iftotalValue
is even) - if you have 2+ entries, count it as an additional pair. For all other numbers, it doesn't matter how many of them you've seen.
$endgroup$
– mabako
Feb 4 at 12:51
add a comment |
$begingroup$
The set approach won't detect 1+1=2
$endgroup$
– Oh My Goodness
Feb 4 at 2:45
$begingroup$
You could very well keep track of how many entries with a value oftotalValue/2
you have (iftotalValue
is even) - if you have 2+ entries, count it as an additional pair. For all other numbers, it doesn't matter how many of them you've seen.
$endgroup$
– mabako
Feb 4 at 12:51
$begingroup$
The set approach won't detect 1+1=2
$endgroup$
– Oh My Goodness
Feb 4 at 2:45
$begingroup$
The set approach won't detect 1+1=2
$endgroup$
– Oh My Goodness
Feb 4 at 2:45
$begingroup$
You could very well keep track of how many entries with a value of
totalValue/2
you have (if totalValue
is even) - if you have 2+ entries, count it as an additional pair. For all other numbers, it doesn't matter how many of them you've seen.$endgroup$
– mabako
Feb 4 at 12:51
$begingroup$
You could very well keep track of how many entries with a value of
totalValue/2
you have (if totalValue
is even) - if you have 2+ entries, count it as an additional pair. For all other numbers, it doesn't matter how many of them you've seen.$endgroup$
– mabako
Feb 4 at 12:51
add a comment |
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1
$begingroup$
Does "unique pairs" mean "unique indexes", "unique values", or "unique combinations of values"? For example: if the data is
1, 1, 1
and the input value is2
, do we have zero, one or three unique pairs?$endgroup$
– Oh My Goodness
Feb 4 at 1:13
1
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@OhMyGoodness, "unique pairs" mean "unique combinations of values" from the list. e.g.: If 'totalValue' we are looking for is 5 and the value at list[i] is 2, list[i+1] is 3 and list[j] 3, list[j+1] is 2 then only one pair (either i, i+1 or j, j+1) should be considered. Hope this helps.
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– silverspoon
Feb 4 at 1:55