I have changed the find function to the filter function and will add the result to the original array.

try this

const arr1 = [
{
  id: 1,
  value: { total: 0 },
},
{
  id: 2,
  value: { total: 0 },
},
{
  id: 3,
  value: { total: 0 },
},
{
  id: 4,
  value: { total: 0 },
}
]
const arr2 = [
    {
      id: 2,
      value: 3 ,
    },
    {
      id: 3,
      value: 5,
    },
  ];
  
arr1.map((item) => {
  let result = arr2.filter((element) => element.id == item.id)
  if(result && result.length) {
    item.value.total += result[0].value
  }
  return item;
});  


console.log(arr1)

You have to assign map to your array:

  this.arr1 = this.arr1.map((item) => {
      this.arr2.find((element) => {
        if (element.id === item.id) {
          item = {

            ...item,
            value: {
              ...item.value,
              total: item.value.total + element.value,
            },
          };
          console.log(item);
        }
      });
      return item;
    });

If each object's id is unique (within each array), you can do:

arr1.forEach((obj1) => {
  const obj2Match = arr2.find((obj2) => obj1.id === obj2.id );

  if (obj2Match) {
    obj1.value.total += obj2Match.value;
  }
});

If more than one object can have the same id (e.g. arr2 has two objects that have the id of 2), then you can do:

arr1.forEach((obj1) => {
  const obj2Matches = arr2.filter((obj2) => obj1.id === obj2.id );

  if (obj2Matches.length) {
    obj1.value.total += obj2Matches.reduce(((accum, curr) => accum  + curr.value), 0);
  }
});

Complexity:

• Time: O(N * M)
• Space: O(M)
• N => arr1.length, M => arr2.length