Place array in alternating positive & negative items with O(1) extra space

Given an array of positive and negative numbers, arrange them in an alternate fashion such that every positive number is followed by negative and vice-versa. Order of elements in output doesn’t matter. Extra positive or negative elements should be moved to end.

Examples:

Input : arr[] = {-2, 3, 4, -1} Output : arr[] = {-2, 3, -1, 4} OR {-1, 3, -2, 4} OR … Input : arr[] = {-2, 3, 1} Output : arr[] = {-2, 3, 1} OR {-2, 1, 3} Input : arr[] = {-5, 3, 4, 5, -6, -2, 8, 9, -1, -4} Output : arr[] = {-5, 3, -2, 5, -6, 4, -4, 9, -1, 8} OR …

Approach 1:

1. First, sort the array in non-increasing order. Then we will count the number of positive and negative integers.
2. Then swap the one negative and one positive number in the odd positions till we reach our condition.
3. This will rearrange the array elements because we are sorting the array and accessing the element from left to right according to our need.

Below is the implementation of the above approach:

• Java

`// Below is the implementation of the above approach`

`import` `java.io.*;`

`import` `java.lang.*;`

`import` `java.util.*;`

`public` `class` `Main {`

` `

` ` `// function which works in the condition when number of`

` ` `// negative numbers are lesser or equal than positive`

` ` `// numbers`

` ` `static` `void` `fill1(` `int` `a[], ` `int` `neg, ` `int` `pos)`

` ` `{`

` ` `if` `(neg % ` `2` `== ` `1` `) {`

` ` `for` `(` `int` `i = ` `1` `; i < neg; i += ` `2` `) {`

` ` `int` `c = a[i];`

` ` `int` `d = a[i + neg];`

` ` `int` `temp = c;`

` ` `a[i] = d;`

` ` `a[i + neg] = temp;`

` ` `}`

` ` `}`

` ` `else` `{`

` ` `for` `(` `int` `i = ` `1` `; i <= neg; i += ` `2` `) {`

` ` `int` `c = a[i];`

` ` `int` `d = a[i + neg - ` `1` `];`

` ` `int` `temp = c;`

` ` `a[i] = d;`

` ` `a[i + neg - ` `1` `] = temp;`

` ` `}`

` ` `}`

` ` `}`

` `

` ` `// Function which works in the condition when number of`

` ` `// negative numbers are greater than positive numbers`

` ` `static` `void` `fill2(` `int` `a[], ` `int` `neg, ` `int` `pos)`

` ` `{`

` ` `if` `(pos % ` `2` `== ` `1` `) {`

` ` `for` `(` `int` `i = ` `1` `; i < pos; i += ` `2` `) {`

` ` `int` `c = a[i];`

` ` `int` `d = a[i + pos];`

` ` `int` `temp = c;`

` ` `a[i] = d;`

` ` `a[i + pos] = temp;`

` ` `}`

` ` `}`

` ` `else` `{`

` ` `for` `(` `int` `i = ` `1` `; i <= pos; i += ` `2` `) {`

` ` `int` `c = a[i];`

` ` `int` `d = a[i + pos - ` `1` `];`

` ` `int` `temp = c;`

` ` `a[i] = d;`

` ` `a[i + pos - ` `1` `] = temp;`

` ` `}`

` ` `}`

` ` `}`

` `

` ` `// Reverse the array`

` ` `static` `void` `reverse(` `int` `a[], ` `int` `n)`

` ` `{`

` ` `int` `i, k, t;`

` ` `for` `(i = ` `0` `; i < n / ` `2` `; i++) {`

` ` `t = a[i];`

` ` `a[i] = a[n - i - ` `1` `];`

` ` `a[n - i - ` `1` `] = t;`

` ` `}`

` ` `}`

` `

` ` `// Print the array`

` ` `static` `void` `print(` `int` `a[], ` `int` `n)`

` ` `{`

` ` `for` `(` `int` `i = ` `0` `; i < n; i++)`

` ` `System.out.print(a[i] + ` `" "` `);`

` ` `System.out.println();`

` ` `}`

` `

` ` `// Driver Code`

` ` `public` `static` `void` `main(String[] args)`

` ` `throws` `java.lang.Exception`

` ` `{`

` ` `// Given array`

` ` `int` `[] arr = { ` `2` `, ` `3` `, -` `4` `, -` `1` `, ` `6` `, -` `9` `};`

` ` `int` `n = arr.length;`

` ` `System.out.println(` `"Given array is "` `);`

` ` `print(arr, n);`

` ` `int` `neg = ` `0` `, pos = ` `0` `;`

` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {`

` ` `if` `(arr[i] < ` `0` `)`

` ` `neg++;`

` ` `else`

` ` `pos++;`

` ` `}`

` ` `// Sort the array`

` ` `Arrays.sort(arr);`

` ` `if` `(neg <= pos) {`

` ` `fill1(arr, neg, pos);`

` ` `}`

` ` `else` `{`

` ` `// reverse the array in this condition`

` ` `reverse(arr, n);`

` ` `fill2(arr, neg, pos);`

` ` `}`

` ` `System.out.println(` `"Rearranged array is "` `);`

` ` `print(arr, n);`

` ` `}`

`}`

Output

Given array is 2 3 -4 -1 6 -9 Rearranged array is -9 3 -1 2 -4 6

Time Complexity: O(N*logN)

Space Complexity: O(1)