Coin Change Dynamic Programming Problem (limited supply of coins)

I'm trying to solve the coin change problem with the variation that the same coin can be used at most twice. I solved the problem where the coins are unlimited but I still don't quite understand how to do it in case of limited coins.

 function coinChange(coins: number[], amount: number): number[][] {

  let M:number[][] = [];
  for (let i = 0 ; i<coins.length ; i++)
    M[i] = [0];

  for(let i = 0 ; i < coins.length ; i++){
    for(let j = 1 ; j <= amount ; j++){
      if(coins[i] > j && i>0) M[i][j] = M[i-1][j];
      else{
        let diff:number = j-coins[i];
        
        if(M[i][diff] != -1){
          let c:number;
          if(i>0){
            c = Math.min(M[i-1][j],1+M[i][diff]);
          }
          else{ 
            c = 1+M[i][diff];
          }
          M[i][j] = c;
        }
        else M[i][j] = -1;
      }
    }
  }
  return M;
}

Solution:

 function coin2(coins:number[], amount:number):number{
  let M:number[][] = [];
  coins.push(...coins)
  for(let i = 0 ; i <= coins.length ; i++){
    M[i] = [];
  }  
  M[0][0] = 0;
  for(let j = 1 ; j <= amount ; j++){
    M[0][j] = Infinity;
  }
  for(let i = 1 ; i <= coins.length ; i++){
    for(let j = 0 ; j <= amount ; j++){
      if(coins[i-1] > j) M[i][j] = M[i-1][j];
      else{
        let dif:number = j-coins[i-1];
        M[i][j] = Math.min(M[i-1][j],1+M[i-1][dif])
      }
    }
  }
  console.log(M);
  return M[coins.length][amount];
}