Need an idea about how to solve this algorithm puzzle
-1
So, it is an C++ assignment question and I have been trying since a long time but couldn't get the idea right. This is the question:
Given two arrays of integers which have the same length, A [0..n-1]
and B [0..n-1]. It is necessary to find the first pair of indices i0
and j0, i0 <= j0, such that A [i0] + B [j0] = max A [i] + B [j], where
0 <= i < n, 0 <= j < n, i <= j.
int maxSum(int arrx, int arry, int x){
int i=0, j=0;
int a;
while(i <= j && j < x){
a = arrx[i] + arry[j];
if(a > arrx[i]){
cout << i << " " << j << " ";
i = x;
}else{
j++;
}
}
return 0;
}
Sample of what should be the I/O:
Input:
4
4 -8 6 0
-10 3 1 1
Output:
0 1
c++
edited Jan 19 at 21:19
melpomene
60.3k54693
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
add a comment |
-1
So, it is an C++ assignment question and I have been trying since a long time but couldn't get the idea right. This is the question:
Given two arrays of integers which have the same length, A [0..n-1]
and B [0..n-1]. It is necessary to find the first pair of indices i0
and j0, i0 <= j0, such that A [i0] + B [j0] = max A [i] + B [j], where
0 <= i < n, 0 <= j < n, i <= j.
int maxSum(int arrx, int arry, int x){
int i=0, j=0;
int a;
while(i <= j && j < x){
a = arrx[i] + arry[j];
if(a > arrx[i]){
cout << i << " " << j << " ";
i = x;
}else{
j++;
}
}
return 0;
}
Sample of what should be the I/O:
Input:
4
4 -8 6 0
-10 3 1 1
Output:
0 1
c++
edited Jan 19 at 21:19
melpomene
60.3k54693
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
And what exactly is your question?
– DYZ
Jan 19 at 21:18
@DYZ It is what is the way to solve such thing?
– Islam A.F.
Jan 19 at 21:20
add a comment |
-1
-1
-1
So, it is an C++ assignment question and I have been trying since a long time but couldn't get the idea right. This is the question:
Given two arrays of integers which have the same length, A [0..n-1]
and B [0..n-1]. It is necessary to find the first pair of indices i0
and j0, i0 <= j0, such that A [i0] + B [j0] = max A [i] + B [j], where
0 <= i < n, 0 <= j < n, i <= j.
int maxSum(int arrx, int arry, int x){
int i=0, j=0;
int a;
while(i <= j && j < x){
a = arrx[i] + arry[j];
if(a > arrx[i]){
cout << i << " " << j << " ";
i = x;
}else{
j++;
}
}
return 0;
}
Sample of what should be the I/O:
Input:
4
4 -8 6 0
-10 3 1 1
Output:
0 1
c++
edited Jan 19 at 21:19
melpomene
60.3k54693
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
So, it is an C++ assignment question and I have been trying since a long time but couldn't get the idea right. This is the question:
Given two arrays of integers which have the same length, A [0..n-1]
and B [0..n-1]. It is necessary to find the first pair of indices i0
and j0, i0 <= j0, such that A [i0] + B [j0] = max A [i] + B [j], where
0 <= i < n, 0 <= j < n, i <= j.
int maxSum(int arrx, int arry, int x){
int i=0, j=0;
int a;
while(i <= j && j < x){
a = arrx[i] + arry[j];
if(a > arrx[i]){
cout << i << " " << j << " ";
i = x;
}else{
j++;
}
}
return 0;
}
Sample of what should be the I/O:
Input:
4
4 -8 6 0
-10 3 1 1
Output:
0 1
c++
c++
edited Jan 19 at 21:19
melpomene
60.3k54693
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
edited Jan 19 at 21:19
melpomene
60.3k54693
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
edited Jan 19 at 21:19
melpomene
60.3k54693
edited Jan 19 at 21:19
melpomene
60.3k54693
edited Jan 19 at 21:19
melpomene
60.3k54693
60.3k54693
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
asked Jan 19 at 21:12
Islam A.F.Islam A.F.
64
64
And what exactly is your question?
– DYZ
Jan 19 at 21:18
@DYZ It is what is the way to solve such thing?
– Islam A.F.
Jan 19 at 21:20
add a comment |
And what exactly is your question?
– DYZ
Jan 19 at 21:18
@DYZ It is what is the way to solve such thing?
– Islam A.F.
Jan 19 at 21:20
And what exactly is your question?
– DYZ
Jan 19 at 21:18
And what exactly is your question?
– DYZ
Jan 19 at 21:18
@DYZ It is what is the way to solve such thing?
– Islam A.F.
Jan 19 at 21:20
@DYZ It is what is the way to solve such thing?
– Islam A.F.
Jan 19 at 21:20
add a comment |
1 Answer
1
active
oldest
votes
0
if I understand your question then, this should work. The answer is 7 that is given from your input data:
int maxSum(int arrx, int arry, int n) // n is the size (count) of the array
{
int i;
int j;
int a;
int maxVal;
int saveI;
int saveJ;
// first, set maxVal to minimum to make sure you get max val (0 is not necessarily min value)
maxVal = 0;
for (i = 0; i <= (n - 1); i++)
{
if (arrx[i] < maxVal)
maxVal = arrx[i];
if (arrx[j] < maxVal)
maxVal = arrx[j];
}
// now, crawl through the arrays
for (j = 0; j <= (n - 1); j++)
{
for (i = 0; i <= j; i++)
{
a = arrx[i] + arry[j];
if (a > maxVal)
{
maxVal = a;
saveI = i;
saveJ = j;
cout << i << " " << j << " " << a " ";
}
}
}
return maxVal;
}
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
Thanks a lot for the effort, appreciated. Worked but had to use my own max value of an array function. But may I know what is the purpose of havingsaveIandsaveJ?
– Islam A.F.
Jan 19 at 22:03
Just incase you wanted to pass them out or access them after the loops have ended.
– IAmNerd2000
Jan 19 at 22:13
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
0
if I understand your question then, this should work. The answer is 7 that is given from your input data:
int maxSum(int arrx, int arry, int n) // n is the size (count) of the array
{
int i;
int j;
int a;
int maxVal;
int saveI;
int saveJ;
// first, set maxVal to minimum to make sure you get max val (0 is not necessarily min value)
maxVal = 0;
for (i = 0; i <= (n - 1); i++)
{
if (arrx[i] < maxVal)
maxVal = arrx[i];
if (arrx[j] < maxVal)
maxVal = arrx[j];
}
// now, crawl through the arrays
for (j = 0; j <= (n - 1); j++)
{
for (i = 0; i <= j; i++)
{
a = arrx[i] + arry[j];
if (a > maxVal)
{
maxVal = a;
saveI = i;
saveJ = j;
cout << i << " " << j << " " << a " ";
}
}
}
return maxVal;
}
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
Thanks a lot for the effort, appreciated. Worked but had to use my own max value of an array function. But may I know what is the purpose of havingsaveIandsaveJ?
– Islam A.F.
Jan 19 at 22:03
Just incase you wanted to pass them out or access them after the loops have ended.
– IAmNerd2000
Jan 19 at 22:13
add a comment |
0
if I understand your question then, this should work. The answer is 7 that is given from your input data:
int maxSum(int arrx, int arry, int n) // n is the size (count) of the array
{
int i;
int j;
int a;
int maxVal;
int saveI;
int saveJ;
// first, set maxVal to minimum to make sure you get max val (0 is not necessarily min value)
maxVal = 0;
for (i = 0; i <= (n - 1); i++)
{
if (arrx[i] < maxVal)
maxVal = arrx[i];
if (arrx[j] < maxVal)
maxVal = arrx[j];
}
// now, crawl through the arrays
for (j = 0; j <= (n - 1); j++)
{
for (i = 0; i <= j; i++)
{
a = arrx[i] + arry[j];
if (a > maxVal)
{
maxVal = a;
saveI = i;
saveJ = j;
cout << i << " " << j << " " << a " ";
}
}
}
return maxVal;
}
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
Thanks a lot for the effort, appreciated. Worked but had to use my own max value of an array function. But may I know what is the purpose of havingsaveIandsaveJ?
– Islam A.F.
Jan 19 at 22:03
Just incase you wanted to pass them out or access them after the loops have ended.
– IAmNerd2000
Jan 19 at 22:13
add a comment |
0
0
0
if I understand your question then, this should work. The answer is 7 that is given from your input data:
int maxSum(int arrx, int arry, int n) // n is the size (count) of the array
{
int i;
int j;
int a;
int maxVal;
int saveI;
int saveJ;
// first, set maxVal to minimum to make sure you get max val (0 is not necessarily min value)
maxVal = 0;
for (i = 0; i <= (n - 1); i++)
{
if (arrx[i] < maxVal)
maxVal = arrx[i];
if (arrx[j] < maxVal)
maxVal = arrx[j];
}
// now, crawl through the arrays
for (j = 0; j <= (n - 1); j++)
{
for (i = 0; i <= j; i++)
{
a = arrx[i] + arry[j];
if (a > maxVal)
{
maxVal = a;
saveI = i;
saveJ = j;
cout << i << " " << j << " " << a " ";
}
}
}
return maxVal;
}
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
if I understand your question then, this should work. The answer is 7 that is given from your input data:
int maxSum(int arrx, int arry, int n) // n is the size (count) of the array
{
int i;
int j;
int a;
int maxVal;
int saveI;
int saveJ;
// first, set maxVal to minimum to make sure you get max val (0 is not necessarily min value)
maxVal = 0;
for (i = 0; i <= (n - 1); i++)
{
if (arrx[i] < maxVal)
maxVal = arrx[i];
if (arrx[j] < maxVal)
maxVal = arrx[j];
}
// now, crawl through the arrays
for (j = 0; j <= (n - 1); j++)
{
for (i = 0; i <= j; i++)
{
a = arrx[i] + arry[j];
if (a > maxVal)
{
maxVal = a;
saveI = i;
saveJ = j;
cout << i << " " << j << " " << a " ";
}
}
}
return maxVal;
}
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
answered Jan 19 at 21:42
IAmNerd2000IAmNerd2000
45419
45419
Thanks a lot for the effort, appreciated. Worked but had to use my own max value of an array function. But may I know what is the purpose of havingsaveIandsaveJ?
– Islam A.F.
Jan 19 at 22:03
Just incase you wanted to pass them out or access them after the loops have ended.
– IAmNerd2000
Jan 19 at 22:13
add a comment |
Thanks a lot for the effort, appreciated. Worked but had to use my own max value of an array function. But may I know what is the purpose of havingsaveIandsaveJ?
– Islam A.F.
Jan 19 at 22:03
Just incase you wanted to pass them out or access them after the loops have ended.
– IAmNerd2000
Jan 19 at 22:13
Thanks a lot for the effort, appreciated. Worked but had to use my own max value of an array function. But may I know what is the purpose of having
saveI and saveJ?– Islam A.F.
Jan 19 at 22:03
Thanks a lot for the effort, appreciated. Worked but had to use my own max value of an array function. But may I know what is the purpose of having
saveI and saveJ?– Islam A.F.
Jan 19 at 22:03
Just incase you wanted to pass them out or access them after the loops have ended.
– IAmNerd2000
Jan 19 at 22:13
Just incase you wanted to pass them out or access them after the loops have ended.
– IAmNerd2000
Jan 19 at 22:13
add a comment |
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And what exactly is your question?
– DYZ
Jan 19 at 21:18
@DYZ It is what is the way to solve such thing?
– Islam A.F.
Jan 19 at 21:20