POTD 1-Sep-2024

04_Array/2D Array/Problem/Easy Problems/2022_Convert_1D_Array_Into_2D_Array.cpp

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+/*
+2022. Convert 1D Array Into 2D Array
+You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with  m rows and n columns using all the elements from original.
+The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on.
+Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.
+Example 1:
+Input: original = [1,2,3,4], m = 2, n = 2
+Output: [[1,2],[3,4]]
+Explanation: The constructed 2D array should contain 2 rows and 2 columns.
+The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
+The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.
+Example 2:
+Input: original = [1,2,3], m = 1, n = 3
+Output: [[1,2,3]]
+Explanation: The constructed 2D array should contain 1 row and 3 columns.
+Put all three elements in original into the first row of the constructed 2D array.
+Example 3:
+Input: original = [1,2], m = 1, n = 1
+Output: []
+Explanation: There are 2 elements in original.
+It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.
+Constraints:
+1 <= original.length <= 5 * 104
+1 <= original[i] <= 105
+1 <= m, n <= 4 * 104
+*/
+class Solution {
+public:
+    vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {
+        if(m*n != original.size()) return {};
+        vector<vector<int>> vec(m,vector<int>(n));
+        int k = 0  ;
+        for(int i = 0 ; i<m && k<original.size();i++)
+            for(int j=0;j<n && k<original.size();j++) vec[i][j] = original[k++];
+    return vec;
+    }
+};

GFG/Medium Problems/Max_sum_path_in_two_arrays.cpp

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+/*
+Max sum path in two arrays
+Given two sorted arrays of distinct integers arr1 and arr2. Each array may have some elements in common with the other array. Find the maximum sum of a path from the beginning of any array to the end of any array. You can switch from one array to another array only at the common elements.
+Note:  When we switch from one array to other,  we need to consider the common element only once in the result.
+Examples : 
+Input: arr1 = [2, 3, 7, 10, 12] , arr2 = [1, 5, 7, 8]
+Output: 35
+Explanation: The path will be 1+5+7+10+12 = 35, where 1 and 5 come from arr2 and then 7 is common so we switch to arr1 and add 10 and 12.
+Input: arr1 = [1, 2, 3] , arr2[] = [3, 4, 5]
+Output: 15
+Explanation: The path will be 1+2+3+4+5=15.
+Expected Time Complexity: O(m + n)
+Expected Auxiliary Space: O(1)
+Constraints:
+1 <= arr1.size(), arr2.size() <= 104
+1 <= arr1[i], arr2[i] <= 105
+*/
+class Solution {
+  public:
+    int maxPathSum(vector<int> &arr1, vector<int> &arr2) {
+        // Code here
+      int i=0;
+        int j=0;
+        int ans=0;
+        int sum1=0;
+        int sum2=0;
+        while(i<arr1.size() && j<arr2.size()){
+            if(arr1[i]>arr2[j]){
+                sum2+=arr2[j];
+                j++;
+            }else if(arr1[i]<arr2[j]){
+                sum1+=arr1[i];
+                i++;
+            }else if(arr1[i]==arr2[j]){
+                ans+=max(sum1,sum2)+arr1[i];
+                sum1=0;
+                sum2=0;
+                i++;
+                j++;
+            }
+        }
+        // ans+=max(sum1,sum2);
+        while(i<arr1.size()){
+            sum1+=arr1[i];
+            i++;
+        }
+        while(j<arr2.size()){
+            sum2+=arr2[j];
+            j++;
+        }
+        ans+=max(sum1,sum2);
+        return ans;
+    }
+};

Leet Code Problems/Easy/2022_Convert_1D_Array_Into_2D_Array.cpp

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+/*
+2022. Convert 1D Array Into 2D Array
+You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with  m rows and n columns using all the elements from original.
+The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on.
+Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.
+Example 1:
+Input: original = [1,2,3,4], m = 2, n = 2
+Output: [[1,2],[3,4]]
+Explanation: The constructed 2D array should contain 2 rows and 2 columns.
+The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
+The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.
+Example 2:
+Input: original = [1,2,3], m = 1, n = 3
+Output: [[1,2,3]]
+Explanation: The constructed 2D array should contain 1 row and 3 columns.
+Put all three elements in original into the first row of the constructed 2D array.
+Example 3:
+Input: original = [1,2], m = 1, n = 1
+Output: []
+Explanation: There are 2 elements in original.
+It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.
+Constraints:
+1 <= original.length <= 5 * 104
+1 <= original[i] <= 105
+1 <= m, n <= 4 * 104
+*/
+class Solution {
+public:
+    vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {
+        if(m*n != original.size()) return {};
+        vector<vector<int>> vec(m,vector<int>(n));
+        int k = 0  ;
+        for(int i = 0 ; i<m && k<original.size();i++)
+            for(int j=0;j<n && k<original.size();j++) vec[i][j] = original[k++];
+    return vec;
+    }
+};