Enhance your Javascript skills by practicing the algorithm to remove negative numbers from an

Question

Enhance your Javascript skills by practicing the algorithm to remove negative numbers from an

I'm really struggling to understand this concept :/

(Eliminating Negative Numbers) Suppose you have an array X containing various values (e.g. [-3,5,1,3,2,10]). Develop a program that removes all negative numbers from the array. By the end of your program, X should solely consist of positive integers. Ensure that you accomplish this task without utilizing a temporary array and only utilize the pop method to eliminate values from the array.

My initial idea was to create a loop through the given array. When X[i] is identified as negative, initiate another loop to swap X[j] and X[j+1] until reaching the end of the array in order to maintain its original order, then use the pop() method.

Upon executing the script, it seems that the loop runs indefinitely. Additionally, if there are two consecutive negative values, the second one might be skipped during the subsequent iteration of i. Is there a more straightforward approach to tackle this problem?

var X = [1,-6,-7,8,9];
//test= [1,-7,8,-6,9]
temp = 0

for (i = 0;i<X.length-1;i++){
    if (X[i]<0){
        for (j = i;j<=X.length-1;j++){
            X[j] = temp
            X[j] = X[j+1] 
            X[j+1] = temp
        }
        if(X[X.length-1] < 0){X.pop()}
    }
};
console.log(X);

javascript arrays algorithm

Answer 1

Answer №1

Another way to tackle the problem is by looping in reverse and swapping negative elements with the last element before popping them off. By removing all negatives from the end first, we can be sure that the last remaining element is positive.

var arr = [1, -6, -7, 8, 9, -3];

// Remove negatives from the end
while (arr.length && arr[arr.length - 1] < 0) {
  arr.pop();
}

for (var i = arr.length - 1; i >= 0; i--) {
  if (arr[i] < 0) {
    // Swap with the last positive element
    arr[i] = arr[arr.length - 1];
    arr.pop();
  }
}

document.body.innerHTML = '<pre>' + JSON.stringify(arr, null, 4) + '</pre>';

This method offers a linear time complexity as it only requires iterating through the list once.

Answer 2

Another way to tackle the problem is by looping in reverse and swapping negative elements with the last element before popping them off. By removing all negatives from the end first, we can be sure that the last remaining element is positive.

var arr = [1, -6, -7, 8, 9, -3];

// Remove negatives from the end
while (arr.length && arr[arr.length - 1] < 0) {
  arr.pop();
}

for (var i = arr.length - 1; i >= 0; i--) {
  if (arr[i] < 0) {
    // Swap with the last positive element
    arr[i] = arr[arr.length - 1];
    arr.pop();
  }
}

document.body.innerHTML = '<pre>' + JSON.stringify(arr, null, 4) + '</pre>';

This method offers a linear time complexity as it only requires iterating through the list once.

Answer 3

Answer №2

Rearrange the array to make sure the negative numbers are positioned at the end.
By implementing a sorting method that shifts the negative values to the tail of the array, we achieve this.

Subsequently, proceed through the elements in reverse order and eliminate those with negative values using the pop function until none remain.

Ultimately, what remains will be solely positive integers.

var X = [-3, 5, 3, 8, 1,-6,-7,8,9];

X.sort(function(a,b) {
    return b - a;
});

for (var i=X.length; i--;) {
    if ( X[i] < 0 ) X.pop();
}

document.body.innerHTML = '<pre>' + JSON.stringify(X, null, 4) + '</pre>';

Answer 4

Rearrange the array to make sure the negative numbers are positioned at the end.
By implementing a sorting method that shifts the negative values to the tail of the array, we achieve this.

Subsequently, proceed through the elements in reverse order and eliminate those with negative values using the pop function until none remain.

Ultimately, what remains will be solely positive integers.

var X = [-3, 5, 3, 8, 1,-6,-7,8,9];

X.sort(function(a,b) {
    return b - a;
});

for (var i=X.length; i--;) {
    if ( X[i] < 0 ) X.pop();
}

document.body.innerHTML = '<pre>' + JSON.stringify(X, null, 4) + '</pre>';

Answer 5

Answer №3

There are already numerous satisfactory responses provided. Here is a simple filter method that avoids sorting the array and utilizes an auxiliary index array j <= i:

function eliminateNegativeValues(arr) {
    var j = 0;

    // filtering the array
    for (var i = 0; i < arr.length; i++) {
        if (arr[i] >= 0) arr[j++] = arr[i];
    }

    // removing surplus elements
    while (j < arr.length) arr.pop();
}

This represents a more traditional C programmer's approach to James Montagne's solution, which is cleaner since it removes elements on-the-fly.

Answer 6

There are already numerous satisfactory responses provided. Here is a simple filter method that avoids sorting the array and utilizes an auxiliary index array j <= i:

function eliminateNegativeValues(arr) {
    var j = 0;

    // filtering the array
    for (var i = 0; i < arr.length; i++) {
        if (arr[i] >= 0) arr[j++] = arr[i];
    }

    // removing surplus elements
    while (j < arr.length) arr.pop();
}

This represents a more traditional C programmer's approach to James Montagne's solution, which is cleaner since it removes elements on-the-fly.

Answer 7

Answer №4

var numbers = [1, -6, -7, 8, 9];
var index = 0;
while (index < numbers.length) {
    if (numbers[index] < 0) {
        numbers[index] = numbers[numbers.length - 1];
        numbers.pop();
    } else {
        index++;
    }
}

only utilizing the pop method on the array

Answer 8

var numbers = [1, -6, -7, 8, 9];
var index = 0;
while (index < numbers.length) {
    if (numbers[index] < 0) {
        numbers[index] = numbers[numbers.length - 1];
        numbers.pop();
    } else {
        index++;
    }
}

only utilizing the pop method on the array

Answer 9

Answer №5

Here is a straightforward approach that eliminates the need for sorting. To achieve this, each element is shifted and pushed if it is not negative. This process repeats itself for the same number of times as the array size. The shifting and pushing can be implemented using either shift/push or pop/unshift.

var originalLength = X.length;

for(var index = 0; index < originalLength; index++) {
  var value = X.pop();

  if(value > 0)
     X.unshift(value);
}

Answer 10

Here is a straightforward approach that eliminates the need for sorting. To achieve this, each element is shifted and pushed if it is not negative. This process repeats itself for the same number of times as the array size. The shifting and pushing can be implemented using either shift/push or pop/unshift.

var originalLength = X.length;

for(var index = 0; index < originalLength; index++) {
  var value = X.pop();

  if(value > 0)
     X.unshift(value);
}

Enhance your Javascript skills by practicing the algorithm to remove negative numbers from an

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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