Approach

Find the leftPrev Node**

  • Create a dummy node to handle edge cases (like when left = 1).
  • Move leftPrev to the node before the left position.

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Reverse the Sublist

  • Use the standard linked list reversal technique within the left to right range.
  • Keep track of prev (previous node) and curr (current node) while reversing.

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Reconnect the Reversed Part

  • The node at leftPrev.next is now the end of the reversed sublist, so connect it to curr.
  • Connect leftPrev.next to prev (the new head of the reversed sublist).

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Code Implementation (JavaScript) 

class ListNode {
    constructor(val = 0, next = null) {
        this.val = val;
        this.next = next;
    }
}

var reverseBetween = function(head, left, right) {
    if (!head || left === right) return head;

    let dummy = new ListNode(0, head);
    let leftPrev = dummy;

    // Step 1: Move leftPrev to the node before 'left'
    for (let i = 1; i < left; i++) {
        leftPrev = leftPrev.next;
    }

    // Step 2: Reverse sublist from 'left' to 'right'
    let prev = null;
    let curr = leftPrev.next;

    for (let i = left; i <= right; i++) {
        let next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }

    // Step 3: Reconnect reversed sublist
    leftPrev.next.next = curr;
    leftPrev.next = prev;

    return dummy.next;
};

Time & Space Complexity

✅ Time Complexity: O(N), since we traverse the list at most twice (finding leftPrev + reversing a segment).

✅ Space Complexity: O(1), since we perform the reversal in-place.