# Delete nodes that have a greater value on the right side

### Problem Statement

Given a singly linked list, implement a function to delete all the nodes which have a greater valued node on the right side.

### Problem Statement Understanding

Let’s understand this problem with the help of some examples. As 15 has a greater value of 20 on its right and similarly, for 14 we have 20 on its right side, so these nodes will be deleted from the linked list.
For all other nodes, we don’t have a greater valued node on its right. So, they will remain in the linked list. If our linked list = 1→2→3→4.

Here, each one of 1,2,3 has some number on the right which is greater than it. So, we remove them.
Only 4 has no greater valued node on its right side, so it will remain in the linked list. I hope these examples helped you in understanding what we need to do.

Now, can you think of an approach to solve this problem?

### Approach 1: Brute force

One straightforward thing we can do is for each node, we traverse all the nodes on the right of it and check whether there is any node having a value greater than it or not. If yes, we simply remove that node from the linked list, else we move to the next node.

We can implement our approach using two nested ‘for loops’.

### Algorithm

1) Iterate over all nodes using a loop
2) For each node
a) Check if there is a node on its right side with a larger value
b) If yes, then delete the current node else, move forward.

### Dry Run ### Code Implementation:

```

#include
using namespace std;

struct Node {
int val;
Node* next;

Node(int value){
val = value;
next = NULL;
}
};

Node* new_node = new Node(new_val);
}

while(i){
cout<val<<" ";
i = i->next;
}
cout<<"\n";
}

int value = i->val;
bool found = false;
for(Node* j = i->next; j!=NULL; j = j->next){
if(j->val > value){
found = 1;
break;
}
}
if(found){
// delete node i
Node* temp = i->next;
i->val = i->next->val;
i->next = i->next->next;
delete temp;
}
else {
i = i->next;
}
}
}

int main(){

// 15 14 20 18 11

// 20 18 11
}
```

#### Output

15 14 20 18 11
20 18 11

Time complexity: O(N*N), due to 2 nested for loops.
Space complexity: O(1), as we don’t use any extra space
Here, N is the total number of nodes in the given linked list.

Can we do some improvements?

If we could traverse from right to left, what improvements would it make?

### Approach 2: Reversing the linked list (optimized)

We can’t traverse a linked list from right to left, let’s reverse it.

In the reversed linked list, we need to do the opposite. We have to remove all the nodes having greater value on the left. Since now, we can traverse from left to right, we can keep track of the maximum value and delete the node whose values are less than the maximum value till that node.

After performing the required deletions, we can again reverse the linked list to get the result in the correct order.

### Algorithm

• Now traverse the linked list and keep track of the maximum value.
• At any node, if the node’s value is smaller than the maximum value till now, delete the node.
• Now we have a linked list in which all nodes having greater value on the left side have been deleted.

### Dry Run ### Code Implementation

```

#include
using namespace std;

struct Node {
int val;
Node* next;

Node(int value){
val = value;
next = NULL;
}
};

Node* new_node = new Node(new_val);
}

while(i){
cout<val<<" ";
i = i->next;
}
cout<<"\n";
}

Node *prev = NULL;
while(curr!=NULL){
next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
return prev;
}

while(curr != NULL){
if(curr->val < mx){
// delete curr node
Node* temp = curr;
prev->next = curr->next;
curr = curr->next;
delete temp;
} else {
mx = max(mx, curr->val);
prev = curr;
curr = curr->next;
}
}
// once again reverse it
}

int main(){

// 15 14 20 18 11