## Linear pass over an array has me stumped

array 3
codingbat arrays

Hi I'm trying to do a single linear pass over two arrays, here is my problem: I have it working with the commented code but it's not correct. I've tried to solve this problem for a whole night before asking for help - here is what I've tried:

////////////////////////////// PROBLEM STATEMENT //////////////////////////////
// Given two arrays of ints sorted in increasing order, outer and inner,     //
// print true if all of the numbers in inner appear in outer. The best       //
// solution makes only a single "linear" pass of both arrays, taking         //
// advantage of the fact that both arrays are already in sorted order.       //
//   {1, 2, 4, 6}, {2, 4} -> true                                            //
//   {1, 2, 4, 6}, {2, 3, 4} -> false                                        //
//   {1, 2, 4, 4, 6}, {2, 4} -> true                                         //
///////////////////////////////////////////////////////////////////////////////

// >>>>>> Your Java Code Fragment starts here <<<<<<
Boolean prez = true;
Scanner kb = new Scanner(System.in);
int n = kb.nextInt(); int z = 0;
int[] inner = new int[n]; for(int i = 0; i < inner.length; i++)inner[i] = kb.nextInt();
n = kb.nextInt();
int[] outer = new int[n]; for(int i = 0; i < outer.length; i++)outer[i] = kb.nextInt();
//   if(outer.length == 0)
//     prez = true;
//  for(int i = 0; i < outer.length; i++){
//    for(int o = 0; o < inner.length ; o++){
//      if(outer[i] != inner[o]){
//        prez = false;
//      } if(outer[i] == inner[o]){
//        prez = true;
//        break;
//      }
//    }
//    if(i == 1 && outer[i] == 3){
//       prez = false;
//       break;
//    }
//  }
//  for(z = 0; z < outer.length ; z++){
//    if(inner[z] == outer[z]){
//      z++;
//      prez = true;
//    }else if(outer[z] != inner[z]){
//      prez = false;
//    }
//  }

int i = 0;
//ok lodo so we loop through the array
for(int j = 0; j < inner.length; j++)
{
//while i is less than the outer length and out[i] is less than inner[j]???
while(i < outer.length && outer[i] < inner[j])
{
//we increase i 1
i++;
}
// if "i" is the same or equal to outer size or the current outer doesnt equal inner make prez false
if(i >= outer.length || outer[i] != inner[j])
{
prez = false;

}
}
//prez = true;
//print the results
System.out.print(prez);

Am i understanding correctly? here is the output

I am trying lodo's solution as about. Here is the output Do i need a break when i get a false?

You need one loop over the inner array, and an external index for the outer array:

boolean checkContained(int[] inner, int[] outer)
{
int i = 0;
for(int j = 0; j < inner.length; j++)
{
while(i < outer.length && outer[i] < inner[j])
{
i++;
}
if(i >= outer.length || outer[i] != inner[j])
{
return false;
}
}
return true;
}

EDIT:

Also (as I see in your output examples), you have to read the outer array first, while in your code you read the inner array first:

int n = kb.nextInt(); int z = 0;
int[] outer= new int[n]; for(int i = 0; i < inner.length; i++)inner[i] = kb.nextInt();
n = kb.nextInt();
int[] inner= new int[n]; for(int i = 0; i < outer.length; i++)outer[i] = kb.nextInt();

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You can manage 2 indexes or 2 iterators, one on each array. A solution with 2 indexes:

boolean ok = true;
int outerIndex = 0;
int innerIndex = 0;
while (ok && outerIndex < outer.length && innerIndex < inner.length) {
if (outer[outerIndex]==inner[innerIndex]) {
outerIndex++;
innerIndex++;
}
else if (outer[outerIndex] < inner[innerIndex]) {
outerIndex++;
}
else {
ok = false;
}
}
// test if there are remaining values in inner array
if (innerIndex < inner.length) {
ok = false;
}
// here ok is true if all inner values have been found in outer

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How would you do this on paper?

You would walk along both array from the start, managing two independent indexes. For the first iteration, both indexes are zero.

Now, you have 3 possibilities:

1. the number at 0 in both arrays are equal
2. the number in the outer array is higher
3. the number in the inner array is higher

So under 1., we can simply increment both the indexes and compare the resulting two numbers.

Under 3., as we know the outer array is sorted, we can increment the index of the outer array and compare the resulting two numbers. In the hopes if finding the value higher up.

Under 2, as we know the inner array is sorted, we know the outer array cannot contain the value and we can abort immediately.

From the logic we can see that we only need one loop, and we only ever increment the indexes, and we always increment at least one index (or exit). So the most we will ever do is travel over both arrays once - making this algorithm linear.

So the skeleton for the code would look like this:

public static boolean containsAll(int[] outer, int[] inner) {
int outerIdx = 0;
int innerIdx = 0;
while (outerIdx < outer.length && innerIdx < inner.length) {
final int o = outer[outerIdx];
final int n = inner[innerIdx];
if (o == n) {
//elements are equal - situation 1.
} else if (o > n) {
//outer element greater than inner - situation 2.
} else if (n > o) {
//inner element greater than outer - situation 3.
}
}
//do we need to check anything here?
}

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Here is my answer hope it explains everything with comments I put in

public boolean linearIn(int[] outer, int[] inner) {

int p = 0;//pointer for inner array

if(inner.length==0)return true;//no inner array

for(int i=0;i<outer.length;i++){//iterating outer array

if(inner[p] == outer[i]){//found a inner array element in outer
p++;//increment pointer of inner array
}
if(p==inner.length){//if everyone in inner array is found
return true;
}
}
return false;
}

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