Find the longest row of consecutive 1's in a matrix in JavaScript-JS Tutorial-php.cn

Find the longest row of consecutive 1's in a matrix in JavaScript

王林

Release： 2023-09-21 17:37:17

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Find the longest row of consecutive 1s in a matrix in JavaScript

Suppose we have a binary matrix (an array containing only arrays of 0 or 1) as shown below -

const arr = [
   [0,1,1,0],
   [0,1,1,0],
   [0,0,0,1]
];

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We need to write a JavaScript function that The function accepts such a matrix as the first and only parameter.

The task of our function is to find the longest row of consecutive matrices in the matrix and return the count of 1's in it. The line can be horizontal, vertical, diagonal or anti-diagonal.

For example, for the above array, the output should be -

const output = 3

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because the longest line starts from arr[0][1] and extends diagonally to -

arr[2][3]

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Example

The code is -

Live demo

const arr = [
   [0,1,1,0],
   [0,1,1,0],
   [0,0,0,1]
];
const longestLine = (arr = []) => {
   if(!arr.length){
      return 0;
   }
   let rows = arr.length, cols = arr[0].length;
   let res = 0;
   const dp = Array(rows).fill([]);
   dp.forEach((el, ind) => {
      dp[ind] = Array(cols).fill([]);
      dp[ind].forEach((undefined, subInd) => {
         dp[ind][subInd] = Array(4).fill(null);
      });
   });
   for (let i = 0; i < rows; i++) {
      for (let j = 0; j < cols; j++) {
         if (arr[i][j] == 1) {
            dp[i][j][0] = j > 0 ? dp[i][j - 1][0] + 1 : 1;
            dp[i][j][1] = i > 0 ? dp[i - 1][j][1] + 1 : 1;
            dp[i][j][2] = (i > 0 && j > 0) ? dp[i - 1][j - 1][2] + 1 : 1;
            dp[i][j][3] = (i > 0 && j < cols - 1) ? dp[i - 1][j + 1][3] + 1 : 1;
            res = Math.max(res, Math.max(dp[i][j][0], dp[i][j][1]));
            res = Math.max(res, Math.max(dp[i][j][2], dp[i][j][3]));
         };
      };
   };
   return res;
};
console.log(longestLine(arr));

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Output

The output in the console will be-

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