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in the hackerrank problem "206. zig zag sequence", the goal is to rearrange a given sequence of integers into a zig-zag pattern. a zig-zag sequence is defined such that, for an array `a`, the following conditions hold:
1. for even indices, the value should be less than the value at the next index (i.e., a[0] a[1], a[2] a[3], ...).
2. for odd indices, the value should be greater than the value at the next index (i.e., a[1] a[2], a[3] a[4], ...).
steps to solve the problem
1. **sort the array**: begin by sorting the array. sorting allows us to easily manage the values needed to create the zig-zag pattern.
2. **rearranging the array**: after sorting, rearrange the array according to the zig-zag conditions. loop through the sorted array and swap elements as necessary to achieve the desired pattern.
3. **output the result**: finally, print or return the rearranged array.
example code
here's a complete code example to illustrate this solution:
explanation of the code
1. **sorting the array**: the `arr.sort()` sorts the input array in ascending order.
2. **looping through the array**: the loop iterates through the array starting from index `1`, incrementing by `2` each time. this means we're looking at the odd indices (1, 3, ...).
3. **swapping elements**: the swap operation `arr[i], arr[i + 1] = arr[i + 1], arr[i]` ensures that the element at the current odd index is greater than the next even index.
4. **returning the result**: after rearranging, the function returns the modified array.
example output
for the input array `[3, 1, 4, 2, 5]`, the sorted array will be `[1, 2, 3, 4, 5]`, and after rearranging, the output will be:
edge cases
1. **single element array**: if the array has only one element, it is already in zig-zag form.
2. **already sorted array**: if the input is already sorted, the function will still rearrange it into a zig-zag pattern.
3. **duplicate elements**: the solution handles duplicates since sorting will group the ...
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