Finding pairs that add up to a given sum, using recursion
0
$begingroup$
I'm currently tasked with using recursion to find pairs that add up to a given sum. How would I make the function recursiveFixedSumPairs function more efficient and truncate this code? Also, what is the current runtime? I believe it is n since it will go through the array only once always starting at the first index.
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findFixedSumPairs(a, k);
}
public static void findFixedSumPairs(int a, int k) {
recursiveFixedSumPairs(a, -1, 0, k);
}
private static void recursiveFixedSumPairs(int array, int subPair1, int index, int k) {
// // pairs whose sum equals k are printed in this method
// if ((array[subPair1] + array[subPair2]) == k) {
// System.out.println("[" + array[subPair1] + ", " + array[subPair2] + "]");
// } else if ((array[subPair1] + array[subPair2]) > k) {
// recursiveFixedSumPairs(array, subPair1, subPair2 - 1, k);
// } else {
// recursiveFixedSumPairs(array, subPair1 + 1, subPair2, k);
// }
if (index == array.length) {
return;
}
else if (index == array.length-1) {
if (subPair1!= -1 && subPair1 + array[index] == k) {
System.out.println(""+subPair1 +" "+ array[index]);
}
else {
return;
}
}
if (subPair1 != -1) {
if (array[index] == k- subPair1) {
System.out.println(""+subPair1 +" "+ array[index]);
return;
} else {
recursiveFixedSumPairs(array, subPair1, index + 1, k);
}
} else {
recursiveFixedSumPairs(array, array[index], index+1, k);
recursiveFixedSumPairs(array,-1, index+1, k);
}
}
}
java recursion complexity k-sum
edited 4 mins ago
200_success
129k15153416
$endgroup$
migrated from stackoverflow.com 2 hours ago
This question came from our site for professional and enthusiast programmers.
add a comment |
0
$begingroup$
I'm currently tasked with using recursion to find pairs that add up to a given sum. How would I make the function recursiveFixedSumPairs function more efficient and truncate this code? Also, what is the current runtime? I believe it is n since it will go through the array only once always starting at the first index.
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findFixedSumPairs(a, k);
}
public static void findFixedSumPairs(int a, int k) {
recursiveFixedSumPairs(a, -1, 0, k);
}
private static void recursiveFixedSumPairs(int array, int subPair1, int index, int k) {
// // pairs whose sum equals k are printed in this method
// if ((array[subPair1] + array[subPair2]) == k) {
// System.out.println("[" + array[subPair1] + ", " + array[subPair2] + "]");
// } else if ((array[subPair1] + array[subPair2]) > k) {
// recursiveFixedSumPairs(array, subPair1, subPair2 - 1, k);
// } else {
// recursiveFixedSumPairs(array, subPair1 + 1, subPair2, k);
// }
if (index == array.length) {
return;
}
else if (index == array.length-1) {
if (subPair1!= -1 && subPair1 + array[index] == k) {
System.out.println(""+subPair1 +" "+ array[index]);
}
else {
return;
}
}
if (subPair1 != -1) {
if (array[index] == k- subPair1) {
System.out.println(""+subPair1 +" "+ array[index]);
return;
} else {
recursiveFixedSumPairs(array, subPair1, index + 1, k);
}
} else {
recursiveFixedSumPairs(array, array[index], index+1, k);
recursiveFixedSumPairs(array,-1, index+1, k);
}
}
}
java recursion complexity k-sum
edited 4 mins ago
200_success
129k15153416
$endgroup$
migrated from stackoverflow.com 2 hours ago
This question came from our site for professional and enthusiast programmers.
2
$begingroup$
Does this code work? If so, Code Review might be a better place to post this.
$endgroup$
– Kevin Anderson
3 hours ago
add a comment |
0
0
0
$begingroup$
I'm currently tasked with using recursion to find pairs that add up to a given sum. How would I make the function recursiveFixedSumPairs function more efficient and truncate this code? Also, what is the current runtime? I believe it is n since it will go through the array only once always starting at the first index.
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findFixedSumPairs(a, k);
}
public static void findFixedSumPairs(int a, int k) {
recursiveFixedSumPairs(a, -1, 0, k);
}
private static void recursiveFixedSumPairs(int array, int subPair1, int index, int k) {
// // pairs whose sum equals k are printed in this method
// if ((array[subPair1] + array[subPair2]) == k) {
// System.out.println("[" + array[subPair1] + ", " + array[subPair2] + "]");
// } else if ((array[subPair1] + array[subPair2]) > k) {
// recursiveFixedSumPairs(array, subPair1, subPair2 - 1, k);
// } else {
// recursiveFixedSumPairs(array, subPair1 + 1, subPair2, k);
// }
if (index == array.length) {
return;
}
else if (index == array.length-1) {
if (subPair1!= -1 && subPair1 + array[index] == k) {
System.out.println(""+subPair1 +" "+ array[index]);
}
else {
return;
}
}
if (subPair1 != -1) {
if (array[index] == k- subPair1) {
System.out.println(""+subPair1 +" "+ array[index]);
return;
} else {
recursiveFixedSumPairs(array, subPair1, index + 1, k);
}
} else {
recursiveFixedSumPairs(array, array[index], index+1, k);
recursiveFixedSumPairs(array,-1, index+1, k);
}
}
}
java recursion complexity k-sum
edited 4 mins ago
200_success
129k15153416
$endgroup$
I'm currently tasked with using recursion to find pairs that add up to a given sum. How would I make the function recursiveFixedSumPairs function more efficient and truncate this code? Also, what is the current runtime? I believe it is n since it will go through the array only once always starting at the first index.
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findFixedSumPairs(a, k);
}
public static void findFixedSumPairs(int a, int k) {
recursiveFixedSumPairs(a, -1, 0, k);
}
private static void recursiveFixedSumPairs(int array, int subPair1, int index, int k) {
// // pairs whose sum equals k are printed in this method
// if ((array[subPair1] + array[subPair2]) == k) {
// System.out.println("[" + array[subPair1] + ", " + array[subPair2] + "]");
// } else if ((array[subPair1] + array[subPair2]) > k) {
// recursiveFixedSumPairs(array, subPair1, subPair2 - 1, k);
// } else {
// recursiveFixedSumPairs(array, subPair1 + 1, subPair2, k);
// }
if (index == array.length) {
return;
}
else if (index == array.length-1) {
if (subPair1!= -1 && subPair1 + array[index] == k) {
System.out.println(""+subPair1 +" "+ array[index]);
}
else {
return;
}
}
if (subPair1 != -1) {
if (array[index] == k- subPair1) {
System.out.println(""+subPair1 +" "+ array[index]);
return;
} else {
recursiveFixedSumPairs(array, subPair1, index + 1, k);
}
} else {
recursiveFixedSumPairs(array, array[index], index+1, k);
recursiveFixedSumPairs(array,-1, index+1, k);
}
}
}
java recursion complexity k-sum
java recursion complexity k-sum
edited 4 mins ago
200_success
129k15153416
edited 4 mins ago
200_success
129k15153416
edited 4 mins ago
200_success
129k15153416
edited 4 mins ago
200_success
129k15153416
edited 4 mins ago
200_success
129k15153416
129k15153416
asked 3 hours ago
FanonX
migrated from stackoverflow.com 2 hours ago
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com 2 hours ago
This question came from our site for professional and enthusiast programmers.
2
$begingroup$
Does this code work? If so, Code Review might be a better place to post this.
$endgroup$
– Kevin Anderson
3 hours ago
add a comment |
2
$begingroup$
Does this code work? If so, Code Review might be a better place to post this.
$endgroup$
– Kevin Anderson
3 hours ago
2
2
$begingroup$
Does this code work? If so, Code Review might be a better place to post this.
$endgroup$
– Kevin Anderson
3 hours ago
$begingroup$
Does this code work? If so, Code Review might be a better place to post this.
$endgroup$
– Kevin Anderson
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
You made it too hard:
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findSumPairs(a, k, 0);
}
private static void findSumPairs(int array, int target, int index) {
for (int i=index+1; i<array.length; i++)
{
if (array[index] + array[i] == target)
{
System.out.println(array[index] + " " + array[i]);
}
}
if (index < array.length -1)
{
findSumPairs(array, target, ++index);
}
}
}
answered 2 hours ago
see sharpersee sharper
101
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
You made it too hard:
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findSumPairs(a, k, 0);
}
private static void findSumPairs(int array, int target, int index) {
for (int i=index+1; i<array.length; i++)
{
if (array[index] + array[i] == target)
{
System.out.println(array[index] + " " + array[i]);
}
}
if (index < array.length -1)
{
findSumPairs(array, target, ++index);
}
}
}
answered 2 hours ago
see sharpersee sharper
101
$endgroup$
add a comment |
0
$begingroup$
You made it too hard:
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findSumPairs(a, k, 0);
}
private static void findSumPairs(int array, int target, int index) {
for (int i=index+1; i<array.length; i++)
{
if (array[index] + array[i] == target)
{
System.out.println(array[index] + " " + array[i]);
}
}
if (index < array.length -1)
{
findSumPairs(array, target, ++index);
}
}
}
answered 2 hours ago
see sharpersee sharper
101
$endgroup$
add a comment |
0
0
0
$begingroup$
You made it too hard:
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findSumPairs(a, k, 0);
}
private static void findSumPairs(int array, int target, int index) {
for (int i=index+1; i<array.length; i++)
{
if (array[index] + array[i] == target)
{
System.out.println(array[index] + " " + array[i]);
}
}
if (index < array.length -1)
{
findSumPairs(array, target, ++index);
}
}
}
answered 2 hours ago
see sharpersee sharper
101
$endgroup$
You made it too hard:
public class HW4 {
public static void main(String args) {
int a = {1, 5, 8, 11, 14, 15, 20, 23, 25, 28, 30, 34};
int k;
k = 43;
System.out.println("k = " + k);
findSumPairs(a, k, 0);
}
private static void findSumPairs(int array, int target, int index) {
for (int i=index+1; i<array.length; i++)
{
if (array[index] + array[i] == target)
{
System.out.println(array[index] + " " + array[i]);
}
}
if (index < array.length -1)
{
findSumPairs(array, target, ++index);
}
}
}
answered 2 hours ago
see sharpersee sharper
101
answered 2 hours ago
see sharpersee sharper
101
answered 2 hours ago
see sharpersee sharper
101
answered 2 hours ago
see sharpersee sharper
101
101
add a comment |
add a comment |
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2
$begingroup$
Does this code work? If so, Code Review might be a better place to post this.
$endgroup$
– Kevin Anderson
3 hours ago