Count Negative Numbers in a Sorted Matrix LeetCode solution in JavaScript

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/**
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 * @param {number[][]} grid
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 * @return {number}
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 */
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var countNegatives = function (grid) {
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  const m = grid.length;
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  const n = grid[0].length;
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  // for each row we need to find location of first negative
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  // all columns past that are negative
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  // also for next row search stops before
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  // location of current first negative because
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  // all numbers below current are also lesser (negative)
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  function binSearch(start, end, arr) {
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    while (start <= end) {
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      let mid = start + ((end - start) >> 1);
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      if (arr[mid] < 0) end = mid - 1;
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      else start = mid + 1;
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    }
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    return arr[end] < 0 ? end : end + 1;
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  }
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  let ans = 0;
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  let searchTill = n - 1;
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  for (let i = 0; i < m; i++) {
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    if (searchTill < 0) {
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      // last row found 0 in first column
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      ans += n;
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      continue;
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    }
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    let firstNegative = binSearch(0, searchTill, grid[i]);
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    searchTill = firstNegative - 1;
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    ans += n - firstNegative;
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  }
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  return ans;
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};