In an interview I was asked the following question. I am given two arrays, both of them are sorted.
BUT
Array 1 will have few -1's and Array 2 will have total numbers as the total number of -1's in Array 1.
So in the below example array1 has three -1's hence array2 has 3 numbers.
let say
var arrayOne = [3,6,-1,11,15,-1,23,34,-1,42];
var arrayTwo = [1,9,28];
Both of the arrays will be sorted.
Now I have to write a program that will merge arrayTwo in arrayOne by replacing -1's, and arrayOne should be in sorted order.
So the output will be
arrayOne = [ 1,3, 6, 9, 11, 15, 23, 28 ,34, 42 ]
Sorting should be done without use of any sort API.
I have written a following code
function puzzle01() {
var arrayOne = [3, 6, -1, 11, 15, -1, 23, 34, -1, 42];
var arrayTwo = [1, 9, 28];
var array1Counter = 0,
isMerged = false;
console.log(" array-1 ", arrayOne);
console.log(" array-2 ", arrayTwo);
for (var array2Counter = 0; array2Counter < arrayTwo.length; array2Counter++) {
isMerged = false;
while (isMerged === false && array1Counter < arrayOne.length) {
if (arrayOne[array1Counter] === -1) {
arrayOne[array1Counter] = arrayTwo[array2Counter];
isMerged = true;
}
array1Counter++;
}
} //for
console.log(" array-1 + array-2 ", arrayOne);
bubbleSort(arrayOne);
console.log(" Sorted array ", arrayOne);
} //puzzle01
puzzle01();
// implementation of bubble sort for sorting the
// merged array
function bubbleSort(arrayOne) {
var nextPointer = 0,
temp = 0,
hasSwapped = false;
do {
hasSwapped = false;
for (var x = 0; x < arrayOne.length; x++) {
nextPointer = x + 1;
if (nextPointer < arrayOne.length && arrayOne[x] > arrayOne[nextPointer]) {
temp = arrayOne[x];
arrayOne[x] = arrayOne[nextPointer];
arrayOne[nextPointer] = temp;
hasSwapped = true;
}
} //for
} while (hasSwapped === true);
} // bubbleSort
The output of the above code is
array-1 [ 3, 6, -1, 11, 15, -1, 23, 34, -1, 42 ]
array-2 [ 1, 9, 28 ]
array-1 + array-2 [ 3, 6, 1, 11, 15, 9, 23, 34, 28, 42 ]
Sorted array [ 1, 3, 6, 9, 11, 15, 23, 28, 34, 42 ]
From the above code you can see, I have first merged the two arrays and than sorted the final one.
Just wanted to know, Is there a better solution.
Is there any flaw in my solution.
Please let me know, it will be helpfull.
After reading all your very helpful comments and answers, I found was able to figure out a more faster solution.
Let us take an example
var arrayOne = [3,6,-1,11,15,-1,32,34,-1,42,-1];
var arrayTwo = [1,10,17,56],
Step1: I will iterate through arrayTwo. Take the next element (i.e. '1') and compare with next element of arrayOne (i.e. '3') and compare.
step 2a : If element of array1 is greater than element of array2 than swap array elements. Now go to next element of array1.
OR
step 2b : If element of array1 is equal to -1 than swap array elements. Now go to next element of array2.
step 3: Go to step 1.
So
in the above example
first iteration, array1 = [1,6,-1,11,...] array2 = [3,10,17,56]
second iteration, array1 = [1,3,-1,11,..] array2 = [6,10,17,56]
third iteration, array1 = [1,3,6,11..] array2 = [-1,10,17,56]
fourth iteration array1 = [1,3,6,10,..] array2 = [-1,11,17,56]
and so on.
at the end I will get the output
array1 = [ 1, 3, 6, 10, 11, 15, 17, 32, 34, 42, 56 ]
array2 = [-1,-1,-1]
Please find the code below,
function puzzle02(arrayOne,arrayTwo){
var array1Counter = 0,
array2Counter = 0,
hasMinusOneOccurred = false;
console.log(" array-1 ",arrayOne);
console.log(" array-2 ",arrayTwo);
while(array2Counter < arrayTwo.length){ // iterate through array2
do{
if(arrayOne[array1Counter] === -1){ // if -1 occurred in array1
hasMinusOneOccurred = true;
// swaping numbers at current position of array1
// with current position of array2
swap(arrayOne,arrayTwo,array1Counter,array2Counter);
// recheck if the current value is greater than other values
// of array1
if(recheckAndSort(arrayOne,array1Counter) === true){
array1Counter = -1;// recheck array1 from start
}else{
// recheck the current array1 counter, for doing so go 1 count back
// so that even if the counter is incremented it points to current
// number itself
array1Counter--;
}
}else if(arrayOne[array1Counter] > arrayTwo[array2Counter]){
swap(arrayOne,arrayTwo,array1Counter,array2Counter);
}else{
array1Counter++;
continue;
}
array1Counter++;
}while(hasMinusOneOccurred === false); // end of do-while
array2Counter++;
hasMinusOneOccurred = false;
}//end of while
console.log(" Sorted array ",arrayOne);
function swap(arr1,arr2,arr1Index,arr2Index){
var temp = arr2[arr2Index];
arr2[arr2Index] = arr1[arr1Index];
arr1[arr1Index] = temp;
}// end of swap
// this method is call if -1 occures in array1
function recheckAndSort(arrayOne,array1Counter){
var isGreaterVal = true,
prevCounter = 0,
nextCounter = 0,
temp = 0,
recheckFromStart = false;
if(array1Counter === 0){ // if -1 occurred at first position of array1.
// flag to check if -1 occurrec at first position
// if yes, iterate array1 from start
recheckFromStart = true;
// iterate forward to check wether any numbers are less than current position,
// if yes than move forward
for(var j = 0; isGreaterVal; j++){
nextCounter = j + 1;
if(arrayOne[nextCounter] === -1){
// swaping numbers of array1 between next to current
swap(arrayOne,arrayOne,nextCounter,j);
isGreaterVal = true;
}else if(arrayOne[nextCounter] < arrayOne[j]){
// swaping numbers of array1 between next to current
swap(arrayOne,arrayOne,nextCounter,j);
isGreaterVal = true;
}else{
isGreaterVal = false;
}
}//end of for
}else{// if -1 occurred in the middle position of array1 and is been swaped then
// iterate backwards to check if any number less then current position exists,
// if yes than shift backwards.
for(var i = array1Counter; isGreaterVal; i--){
prevCounter = i - 1;
if(arrayOne[prevCounter] > arrayOne[i]){
// swaping numbers of array1 between previous to current
swap(arrayOne,arrayOne,prevCounter,i);
isGreaterVal = true;
}else{
isGreaterVal = false;
}
}// end of for
}
return recheckFromStart;
}// end of recheckAndSort
} // end of puzzle02
After calling the above function
puzzle02([3,6,-1,11,15,-1,32,34,-1,42,-1],[1,10,17,56]);
The output of above code is,
array-1 [ 3, 6, -1, 11, 15, -1, 32, 34, -1, 42, -1 ]
array-2 [ 1, 10, 17, 56 ]
Sorted array [ 1, 3, 6, 10, 11, 15, 17, 32, 34, 42, 56 ]
Thanks.
You can try following approach:
Logic
- Create a new array that will be returned.
- Check for first element in
arrayTwoand keep it in a variable sayval. - Loop over
arrayOneand check if current value is greater thanval, push it in array and decrement value ofiby 1 to check next value as well with current element. - Now check for current element. If it is less than
0, ignore it, else push value to array. - Return this array.
function mergeAndSort(a1, a2) {
var matchCount = 0;
var ret = [];
for (var i = 0; i < a1.length; i++) {
var val = a2[matchCount];
if (a1[i] > val) {
ret.push(val)
matchCount++
i--;
continue;
}
if (a1[i] > 0) {
ret.push(a1[i]);
}
}
console.log(ret.join())
return ret;
}
var arrayOne = [3, 6, -1, 11, 15, -1, 23, 34, -1, 42]
var arrayTwo = [7, 19, 38];
var arrayThree = [1, 9, 28];
var arrayFour = [1,2,5]
mergeAndSort(arrayOne, arrayTwo)
mergeAndSort(arrayOne, arrayThree)
mergeAndSort(arrayOne, arrayFour)
.as-console-wrapper {
max-height: 100% !important;
top: 0;
}
Note: Not putting check for number of elements in
arrayTwo as its clearly mentioned in question that it will be same.
Posts
![IVRAMANA](https://draft.blogger.com/img/profile_blank_gray.jpg" title="IVRAMANA)
0 comments:
Post a Comment