<p>Given the <code>head</code> of a linked list, we repeatedly delete consecutive sequences of nodes that sum to <code>0</code> until there are no such sequences.</p>

<p>After doing so, return the head of the final linked list.&nbsp; You may return any such answer.</p>

<p>&nbsp;</p>
<p>(Note that in the examples below, all sequences are serializations of <code>ListNode</code> objects.)</p>

<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,-3,3,1]
<strong>Output:</strong> [3,1]
<strong>Note:</strong> The answer [1,2,1] would also be accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,3,-3,4]
<strong>Output:</strong> [1,2,4]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,3,-3,-2]
<strong>Output:</strong> [1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The given linked list will contain between <code>1</code> and <code>1000</code> nodes.</li>
	<li>Each node in the linked list has <code>-1000 &lt;= node.val &lt;= 1000</code>.</li>
</ul>

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeZeroSumSublists(ListNode* head) {
        vector<int> prefixSum = {0} ; 
        vector<int> nodes ;
        for(ListNode * curr = head ; curr != nullptr ; curr= curr->next ) {
            prefixSum.push_back(prefixSum.back()+curr->val) ; 
            nodes.push_back(curr->val) ; 
        }
        unordered_map<int,int> mp ; 
        for(int i = 0 ; i<prefixSum.size() ; i++){
            if(mp.find(prefixSum[i]) != mp.end()){
                int k = mp[prefixSum[i]] ; 
                while(k<i){
                    mp.erase(prefixSum[k]) ; 
                    nodes[k] = INT_MIN ; 
                    k++ ; 
                }
            }
            mp[prefixSum[i]] = i ; 
        }
        ListNode * dummyHead = new ListNode() ; 
        ListNode * newCurr = dummyHead ; 
        for(int i = 0 ; i<nodes.size() ; i++){
            if(nodes[i]!= INT_MIN){
            newCurr->next = new ListNode(nodes[i]) ; 
            newCurr = newCurr ->next ; }
        }
        return dummyHead->next ; 
    }
};
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