/* * Given an array of integers and a number X, * find the longest non-empty subarray with a minimum equal to X. * Return the length of such a subarray or -1 if none exists. */

Task: find the longest non-empty subarray of the array where the minimum equals X. If there are no such subarrays, return -1.

Solution idea:

• Iterate through the array, splitting it into subarrays where the minimum equals X.
• The minimum of a subarray is its smallest element. For the minimum to be X, the subarray must not contain elements less than X, and must contain at least one element equal to X.

Algorithm:

1. Traverse the array, dividing it into segments where all elements >= X.
2. For each such segment, check if it contains an element equal to X.
3. If it does, update the maximum length.

Example implementation in Go:

func longestSubarrayWithMinX(arr []int, X int) int {
    maxLen := -1
    start := 0
    hasX := false

    for i, val := range arr {
        if val < X {
            // End the current segment
            if hasX && i-start > maxLen {
                maxLen = i - start
            }
            start = i + 1
            hasX = false
        } else {
            if val == X {
                hasX = true
            }
        }
    }

    // Check the last segment
    if hasX && len(arr)-start > maxLen {
        maxLen = len(arr) - start
    }

    return maxLen
}

This code traverses the array once, maintaining the current segment with elements >= X and a flag indicating the presence of X inside it. If an element < X is encountered, the segment ends. In the end, it returns the maximum length of a suitable subarray or -1 if none exists.