hashmap for O{1) lookup, maintain row and col arrays for count of uncolored numbers

2661. First Completely Painted Row or Column
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Medium
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Hint
You are given a 0-indexed integer array arr, and an m x n integer matrix mat. arr and mat both contain all the integers in the range [1, m * n].

Go through each index i in arr starting from index 0 and paint the cell in mat containing the integer arr[i].

Return the smallest index i at which either a row or a column will be completely painted in mat.

 

Example 1:

image explanation for example 1
Input: arr = [1,3,4,2], mat = [[1,4],[2,3]]
Output: 2
Explanation: The moves are shown in order, and both the first row and second column of the matrix become fully painted at arr[2].
Example 2:

image explanation for example 2
Input: arr = [2,8,7,4,1,3,5,6,9], mat = [[3,2,5],[1,4,6],[8,7,9]]
Output: 3
Explanation: The second column becomes fully painted at arr[3].
 

Constraints:

m == mat.length
n = mat[i].length
arr.length == m * n
1 <= m, n <= 105
1 <= m * n <= 105
1 <= arr[i], mat[r][c] <= m * n
All the integers of arr are unique.
All the integers of mat are unique.

Seen this question in a real interview before?
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Accepted
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Acceptance Rate
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Topics
Array
Hash Table
Matrix
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Hint 1
Can we use a frequency array?
Hint 2
Pre-process the positions of the values in the matrix.
Hint 3
Traverse the array and increment the corresponding row and column frequency using the pre-processed positions.
Hint 4
If the row frequency becomes equal to the number of columns, or vice-versa return the current index.

Author: Gauranshgoel123

first-completely-painted-row-or-column.java

@@ -0,0 +1,47 @@
+class Solution {
+    public int firstCompleteIndex(int[] arr, int[][] mat) {
+        int row=mat.length;
+        int col=mat[0].length;
+        Map<Integer,int[]> map=new HashMap<>();
+        for(int i=0;i<row;i++){
+            for(int j=0;j<col;j++){
+                map.put(mat[i][j],new int[]{i,j});
+            }
+        }
+        int[] rowcount=new int[row];
+        int[] colcount=new int[col];
+        Arrays.fill(rowcount,col);
+        Arrays.fill(colcount,row);
+        
+        for(int i=0;i<arr.length;i++){
+            int[] pos=map.get(arr[i]);
+            if(--rowcount[pos[0]]==0 || --colcount[pos[1]]==0) return i;
+        }
+        return -1;
+    }
+}
+
+
+
+/*
+mat = [[3,2,5],[1,4,6],[8,7,9]]
+
+arr = [2,8,7,4,1,3,5,6,9], 
+       0 1 2 3 4 5 6 7 8
+
+first three computed normally
+       2 done
+       8 done
+       7 done
+
+       check now onwards if on computing any row or coloumn
+       is formed
+
+
+
+
+
+
+
+
+*/