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  1. Practice
  2. Data Structures
  3. Arrays
  4. Array Manipulation
  5. Discussions

Array Manipulation

  • krishnna + 1 comment

    check it out here, you will get all your doubts solved https://www.geeksforgeeks.org/difference-array-range-update-query-o1/

    • tejaspatil + 1 comment

      Below link will also help to understand theory behind it. https://www.geeksforgeeks.org/constant-time-range-add-operation-array/

      • codam + 4 comments

        Same solution in Javascript

            var arr = [];
    var max = 0;
    // init each element of arr to 0
    for (let l = 0; l < n; l++) {
        arr[l] = 0;
    }
    // for each sum operation in queries
    for (let i = 0; i < queries.length; i++) {
       // update arr with number to add at index=queries[i][0]  and number to remove at index=queries[i][0]+1 => this will allow us to build each element of the final array by summing all elements before it. The aim of this trick is to lower time complexity
        arr[queries[i][0]-1] += queries[i][2];
        if (queries[i][1] < arr.length) {
            arr[queries[i][1]] -= queries[i][2];
        }
    }
    for (let j = 1; j < n; j++) {
        arr[j] += arr[j-1];
    }
    for (let k = 0; k < arr.length; k++) {
        max = Math.max(max, arr[k]);
    }
    //max = Math.max(...arr); // not working for big arrays
    return max;
        • kamnijais + 1 comment

          Hey, I did the code in Java8 and my code is getting failed for input type - where only single value is present in a row of array. meaning only left index value is provided and right and k value is missing from array. So can you help me how to solve this issue?

          • h1017691668 + 0 comments

            you could post your code,and we can check it out

        • mevtew + 1 comment

          simpler in es6:

          function arrayManipulation(n, queries) { let arr = new Array(2*n).fill(0); let max = 0;

          queries.forEach((item) => {
    arr[item[0]] += item[2];
    arr[item[1] + 1] -= item[2];
});

arr.reduce((prev, curr, idx) => {
    const sum = prev + curr;
    if (sum > max) {
        max = sum;
    }

    return sum;
})

return max;

          }

          • Rockyfive + 0 comments

            I did something pretty similar, just with a little bit more readable forEach:

            const arr = new Array(n).fill(0);
let result = 0;

queries.forEach(([a, b, k]) => {
    arr[a - 1] += k;
    if (b < arr.length) {
        arr[b] -= k;
    }
});

arr.reduce((a, b) => {
    const acc = a + b;
    result = Math.max(result, acc);
    return acc;
}, 0);

return result;
        • jpeavey451 + 0 comments

          Your last for loop isn't needed. You can move Math.max to the previous for loop.

        • lyons_mr + 0 comments
          [deleted]