判断A、B数组的包含关系（值和数量），A属于B返回1，B属于A返回2，两者相等返回0，其他返回-1

[Sunny Fu]

[beary]

const A = [1, 3, 5, 6, 7];
const B = [1, 3, 5, 6, 8];
function judge(A, B) {
  let str1 = A.sort((a, b) => a - b).join("");
  let str2 = B.sort((a, b) => a - b).join("");
  if(str1 === str2) return 0;
  else if (str1.includes(str2)) return 2;
  else if (str2.includes(str1)) return 1;
  return -1;
}
console.log(judge(A, B));

[kangkang123269]

function arrayToFrequencyObject(arr) {
    return arr.reduce((obj, item) => {
        obj[item] = (obj[item] || 0) + 1;
        return obj;
    }, {});
}

function compareArrays(A, B) {
    const freqA = arrayToFrequencyObject(A);
    const freqB = arrayToFrequencyObject(B);

    let isASubsetOfB = true;
    let isBSubsetOfA = true;

    for (let key in freqA) {
        if (!(key in freqB && freqA[key] <= freqB[key])) {
            isASubsetOfB = false;
            break;
        }
    }

    for (let key in freqB) {
        if (!(key in freqA && freqB[key] <= freqA[key])) {
            isBSubsetOfA = false;
            break;
        }
    }

   if(isASubsetOfB && isBSubsetOfA){
       return 0; // A and B are equal
   } else if(isASubsetOfB){
       return 1; // A is a subset of B
   } else if(isBSubsetOfA){
       return 2; // B is a subset of A
   } else{
       return -1; // A and B are disjoint
   }
}

[gongshiwei]

function judge(arrA, arrB) {
    let key = true
    for (const item of arrA) {
        if (!arrB.includes(item)) {
            key = false
            break;
        }
    }
    if (key) {
        if (arrA.length === arrB.length) return 0
        else return 1
    } else {
        key = true
        for (const item of arrB) {
            if (!arrA.includes(item)) {
                key = false
                break;
            }
        }
        if (key) return 2
    }
    return -1
}