Find two numbers with target sum
$begingroup$
I am trying to solve this problem:
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
and this is my implementation:
public int twoSum(int numbers, int target) {
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>();
int requiredNumbers = null;
int index = 0;
for (int number : numbers) {
if (numbersMap.containsKey(target - number)) {
requiredNumbers = new int[2];
requiredNumbers[0] = numbersMap.get(target - number);
requiredNumbers[1] = index;
return requiredNumbers;
} else {
numbersMap.put(number, index);
index++;
}
}
return requiredNumbers;
}
How can I improve its execution time?
java performance programming-challenge
edited Dec 23 '18 at 18:30
Ankit Soni
483111
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
$endgroup$
bumped to the homepage by Community♦ 7 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I am trying to solve this problem:
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
and this is my implementation:
public int twoSum(int numbers, int target) {
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>();
int requiredNumbers = null;
int index = 0;
for (int number : numbers) {
if (numbersMap.containsKey(target - number)) {
requiredNumbers = new int[2];
requiredNumbers[0] = numbersMap.get(target - number);
requiredNumbers[1] = index;
return requiredNumbers;
} else {
numbersMap.put(number, index);
index++;
}
}
return requiredNumbers;
}
How can I improve its execution time?
java performance programming-challenge
edited Dec 23 '18 at 18:30
Ankit Soni
483111
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
$endgroup$
bumped to the homepage by Community♦ 7 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
$begingroup$
you could have used indexed for loop as you are anyway keeping the track of the index
$endgroup$
– Ankit Soni
Dec 23 '18 at 18:24
add a comment |
$begingroup$
I am trying to solve this problem:
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
and this is my implementation:
public int twoSum(int numbers, int target) {
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>();
int requiredNumbers = null;
int index = 0;
for (int number : numbers) {
if (numbersMap.containsKey(target - number)) {
requiredNumbers = new int[2];
requiredNumbers[0] = numbersMap.get(target - number);
requiredNumbers[1] = index;
return requiredNumbers;
} else {
numbersMap.put(number, index);
index++;
}
}
return requiredNumbers;
}
How can I improve its execution time?
java performance programming-challenge
edited Dec 23 '18 at 18:30
Ankit Soni
483111
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
$endgroup$
I am trying to solve this problem:
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
and this is my implementation:
public int twoSum(int numbers, int target) {
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>();
int requiredNumbers = null;
int index = 0;
for (int number : numbers) {
if (numbersMap.containsKey(target - number)) {
requiredNumbers = new int[2];
requiredNumbers[0] = numbersMap.get(target - number);
requiredNumbers[1] = index;
return requiredNumbers;
} else {
numbersMap.put(number, index);
index++;
}
}
return requiredNumbers;
}
How can I improve its execution time?
java performance programming-challenge
java performance programming-challenge
edited Dec 23 '18 at 18:30
Ankit Soni
483111
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
edited Dec 23 '18 at 18:30
Ankit Soni
483111
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
edited Dec 23 '18 at 18:30
Ankit Soni
483111
edited Dec 23 '18 at 18:30
Ankit Soni
483111
edited Dec 23 '18 at 18:30
Ankit Soni
483111
483111
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
asked Dec 23 '18 at 15:12
Karan KhannaKaran Khanna
1656
1656
bumped to the homepage by Community♦ 7 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 7 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
$begingroup$
you could have used indexed for loop as you are anyway keeping the track of the index
$endgroup$
– Ankit Soni
Dec 23 '18 at 18:24
add a comment |
1
$begingroup$
you could have used indexed for loop as you are anyway keeping the track of the index
$endgroup$
– Ankit Soni
Dec 23 '18 at 18:24
1
1
$begingroup$
you could have used indexed for loop as you are anyway keeping the track of the index
$endgroup$
– Ankit Soni
Dec 23 '18 at 18:24
$begingroup$
you could have used indexed for loop as you are anyway keeping the track of the index
$endgroup$
– Ankit Soni
Dec 23 '18 at 18:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the size of your input array can be large, you can get a speed-up by preallocating the capacity of your HashMap:
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>(numbers.length * 2);
As the algorithm runs, data will be added to the HashMap. When number of entries exceeds the capacity * load_factor, the hashmap's capacity is doubled, and the elements are re-binned for the larger capacity. This capacity doubling and rebinning takes time. It doesn't happen often, $O(log N)$ times, but it can be eliminated by starting with a hashmap of sufficient capacity.
The load_factor defaults to 0.75, so an initial capacity larger than numbers.length * 4/3 is required. numbers.length * 2 is a simple expression that satisfies that requirement.
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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oldest
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votes
$begingroup$
If the size of your input array can be large, you can get a speed-up by preallocating the capacity of your HashMap:
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>(numbers.length * 2);
As the algorithm runs, data will be added to the HashMap. When number of entries exceeds the capacity * load_factor, the hashmap's capacity is doubled, and the elements are re-binned for the larger capacity. This capacity doubling and rebinning takes time. It doesn't happen often, $O(log N)$ times, but it can be eliminated by starting with a hashmap of sufficient capacity.
The load_factor defaults to 0.75, so an initial capacity larger than numbers.length * 4/3 is required. numbers.length * 2 is a simple expression that satisfies that requirement.
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
$endgroup$
add a comment |
$begingroup$
If the size of your input array can be large, you can get a speed-up by preallocating the capacity of your HashMap:
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>(numbers.length * 2);
As the algorithm runs, data will be added to the HashMap. When number of entries exceeds the capacity * load_factor, the hashmap's capacity is doubled, and the elements are re-binned for the larger capacity. This capacity doubling and rebinning takes time. It doesn't happen often, $O(log N)$ times, but it can be eliminated by starting with a hashmap of sufficient capacity.
The load_factor defaults to 0.75, so an initial capacity larger than numbers.length * 4/3 is required. numbers.length * 2 is a simple expression that satisfies that requirement.
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
$endgroup$
add a comment |
$begingroup$
If the size of your input array can be large, you can get a speed-up by preallocating the capacity of your HashMap:
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>(numbers.length * 2);
As the algorithm runs, data will be added to the HashMap. When number of entries exceeds the capacity * load_factor, the hashmap's capacity is doubled, and the elements are re-binned for the larger capacity. This capacity doubling and rebinning takes time. It doesn't happen often, $O(log N)$ times, but it can be eliminated by starting with a hashmap of sufficient capacity.
The load_factor defaults to 0.75, so an initial capacity larger than numbers.length * 4/3 is required. numbers.length * 2 is a simple expression that satisfies that requirement.
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
$endgroup$
If the size of your input array can be large, you can get a speed-up by preallocating the capacity of your HashMap:
Map<Integer, Integer> numbersMap = new HashMap<Integer, Integer>(numbers.length * 2);
As the algorithm runs, data will be added to the HashMap. When number of entries exceeds the capacity * load_factor, the hashmap's capacity is doubled, and the elements are re-binned for the larger capacity. This capacity doubling and rebinning takes time. It doesn't happen often, $O(log N)$ times, but it can be eliminated by starting with a hashmap of sufficient capacity.
The load_factor defaults to 0.75, so an initial capacity larger than numbers.length * 4/3 is required. numbers.length * 2 is a simple expression that satisfies that requirement.
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
answered Dec 23 '18 at 19:55
AJNeufeldAJNeufeld
4,497318
4,497318
add a comment |
add a comment |
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1
$begingroup$
you could have used indexed for loop as you are anyway keeping the track of the index
$endgroup$
– Ankit Soni
Dec 23 '18 at 18:24