Minimum Falling Path Sum II LeetCode solution in JavaScript

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var minFallingPathSum = function (grid) {
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  if (grid.length === 1) {
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    return grid[0][0];
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  }
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  let sums = []; //find first, second and third minimum values in the grid as these are only possible values for path
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  //store the index for fast lookup in a 3D array
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  for (let i = 0; i < grid.length; i++) {
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    let min = Math.min(...grid[i]);
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    let idx = grid[i].indexOf(min);
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    grid[i].splice(idx, 1, Infinity);
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    let min2 = Math.min(...grid[i]);
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    let idx2 = grid[i].indexOf(min2);
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    grid[i].splice(idx2, 1, Infinity);
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    let min3 = Math.min(...grid[i]);
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    let idx3 = grid[i].indexOf(min3);
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    sums.push([
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      [min, idx],
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      [min2, idx2],
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      [min3, idx3],
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    ]);
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  }
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  // sums contains 1st , 2nd and 3rd min value and index
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  let memo = {}; // for memoization
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  let recur = (r, li) => {
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    //recursive call to decide which path to choose from r=row, li=lastIndex
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    if (memo[r + "," + li] !== undefined) {
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      // memoized
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      return memo[r + "," + li];
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    }
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    if (r === grid.length - 1) {
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      // last and first choice for path can only be from 1st and 2nd minimum, as there is only one other row to occupy these columns
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      if (li === sums[r][0][1]) {
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        memo[r + "," + li] = sums[r][1][0];
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      } else {
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        memo[r + "," + li] = sums[r][0][0];
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      }
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      return memo[r + "," + li];
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    } else {
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      switch (
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        li // pick the minimum value path where index is not equal to last index
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      ) {
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        case sums[r][0][1]:
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          memo[r + "," + li] = Math.min(
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            sums[r][1][0] + recur(r + 1, sums[r][1][1]),
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            sums[r][2][0] + recur(r + 1, sums[r][2][1])
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          );
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          break;
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        case sums[r][1][1]:
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          memo[r + "," + li] = Math.min(
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            sums[r][0][0] + recur(r + 1, sums[r][0][1]),
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            sums[r][2][0] + recur(r + 1, sums[r][2][1])
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          );
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          break;
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        case sums[r][2][1]:
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          memo[r + "," + li] = Math.min(
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            sums[r][1][0] + recur(r + 1, sums[r][1][1]),
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            sums[r][0][0] + recur(r + 1, sums[r][0][1])
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          );
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          break;
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        default:
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          memo[r + "," + li] = Math.min(
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            sums[r][0][0] + recur(r + 1, sums[r][0][1]),
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            sums[r][1][0] + recur(r + 1, sums[r][1][1]),
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            sums[r][2][0] + recur(r + 1, sums[r][2][1])
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          );
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      }
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      return memo[r + "," + li];
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    }
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  };
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  return recur(0, sums[0][2][1]); //since first and last can only be picked from 1st and 2nd min values
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};