Added solution to lc-846

Author: nagalakshmi08

dsa-solutions/lc-solutions/0800- 0899/0846-hand-of-straights.md

@@ -0,0 +1,187 @@
+---
+id: hand-of-straights
+title: 846. Hand of Straights
+sidebar_label: 846. Hand of Straights
+tags:
+- Array
+- Hash Table
+- Greedy
+- Sorting
+
+description: "This is a solution to the 846. Hand of Straights."
+---
+
+## Problem Description
+Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size groupSize, and consists of groupSize consecutive cards.
+
+Given an integer array hand where hand[i] is the value written on the ith card and an integer groupSize, return true if she can rearrange the cards, or false otherwise.
+
+ ### Examples
+**Example 1:**
+```
+Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3
+
+Output: true
+
+Explanation:
+Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8]
+```
+
+**Example 2:**
+```
+Input: [1,2,3,4,5], groupSize = 4
+
+Output: false
+
+Explanation:
+Alice's hand can not be rearranged into groups of 4.
+```
+
+### Constraints
+- `1 <= hand.length <= 10^4`
+- `0 <= hand[i] <= 10^9`
+- `1 <= groupSize <= hand.length`
+## Solution for 846. Hand of Straights
+
+The problem is about arranging cards into groups of consecutive sequences. The key observation is that, to form valid groups, each smallest number in the group should lead to groupSize - 1 consecutive numbers following it. If we can always find such sequences starting from the smallest number, then it is possible to arrange the cards into the desired groups.
+
+## Approach
+
+1. **Sort the Cards**: The first step is to sort the cards. Sorting helps us to easily identify and form consecutive sequences.
+
+2. **Find Successors**: For each number in the sorted array, if it hasn't already been used in a group (indicated by a value of -1), try to form a group starting with that number. Use a helper function findSuccessors to check if it is possible to form a valid group starting from the current number.
+
+3. **Mark Used Cards**: Once a card is used in a group, mark it as `-1` to indicate it has been processed.
+
+4. **Check Group Formation**: For each number, use the helper function to see if a group of groupSize consecutive numbers can be formed. If at any point, it is not possible to form such a group, return false.
+
+5. **Completion Check**: If all numbers can be grouped correctly, return true.
+
+### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@nagalakshmi08"/>
+
+```cpp
+class Solution {
+public:
+   bool findSuccessors(vector<int>& hand, int groupSize, int i, int n) {
+        int next = hand[i] + 1;
+        hand[i] = -1;  // Mark as used
+        int count = 1;
+        i += 1;
+        while (i < n && count < groupSize) {
+            if (hand[i] == next) {
+                next = hand[i] + 1;
+                hand[i] = -1;
+                count++;
+            }
+            i++;
+        }
+        return count == groupSize;
+    }
+
+    bool isNStraightHand(vector<int>& hand, int groupSize) {
+        int n = hand.size();
+        if (n % groupSize != 0) return false;
+        std::sort(hand.begin(), hand.end());
+        for (int i = 0; i < n; i++) {
+            if (hand[i] >= 0) {
+                if (!findSuccessors(hand, groupSize, i, n)) return false;
+            }
+        }
+        return true;
+    }
+};
+```
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@nagalakshmi08"/>
+```java
+class Solution {
+    public boolean findsucessors(int[] hand, int groupSize, int i, int n) {
+        int f = hand[i] + 1;
+        hand[i] = -1;
+        int count = 1;
+        i += 1;
+        while (i < n && count < groupSize) {
+            if (hand[i] == f) {
+                f = hand[i] + 1;
+                hand[i] = -1;
+                count++;
+            }
+            i++;
+        }
+        if (count != groupSize)
+            return false;
+        else
+            return true;
+    }
+
+    public boolean isNStraightHand(int[] hand, int groupSize) {
+        int n = hand.length;
+        if (n % groupSize != 0)
+            return false;
+        Arrays.sort(hand);
+        int i = 0;
+        for (; i < n; i++) {
+            if (hand[i] >= 0) {
+                if (!findsucessors(hand, groupSize, i, n))
+                    return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
+</TabItem>
+
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@nagalakshmi08"/>
+
+```python
+class Solution(object):
+    def find_successors(self, hand, groupSize, i, n):
+        next_val = hand[i] + 1
+        hand[i] = -1  # Mark as used
+        count = 1
+        i += 1
+        while i < n and count < groupSize:
+            if hand[i] == next_val:
+                next_val = hand[i] + 1
+                hand[i] = -1
+                count += 1
+            i += 1
+        return count == groupSize
+
+    def isNStraightHand(self, hand, groupSize):
+        n = len(hand)
+        if n % groupSize != 0:
+            return False
+        hand.sort()
+        for i in range(n):
+            if hand[i] >= 0:
+                if not self.find_successors(hand, groupSize, i, n):
+                    return False
+        return True
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- **Time Complexity**: Sorting the array takes $(O(n \log n))$. The subsequent grouping operation in the worst case can take $(O(n \times groupSize))$, leading to an overall time complexity of $(O(n \log n + n \times groupSize))$.
+- **Space Complexity**: IThe space complexity is $(O(1))$ for the in-place operations (aside from the input array).
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['nagalakshmi08'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>