I have this usecase that I thought will be easy to achieve, but I stuck. There is a Row with 3 elements of different width. I want to place them in the same distance, but distance should be measured from the center of the object, not the gap between them. Here is the screenshot of standard Row with Arrangement.SpaceBetween. How can I position them , so the x1 == x2 ?

You can write your own

Arrangement

that would do what you need. It’s not hard. This looks like

justified

arrangement I wrote recently. It’s not perfect for RTL/LTR and maybe something else, but it could give you an idea

fun Arrangement.justified(minSpace: Dp): Arrangement.HorizontalOrVertical = object : Arrangement.HorizontalOrVertical {
    override val spacing: Dp
        get() = minSpace

    override fun Density.arrange(totalSize: Int, sizes: IntArray, layoutDirection: LayoutDirection, outPositions: IntArray) {
        if (sizes.isEmpty()) return
        val minSpacePx = minSpace.roundToPx()
        val measuredSize = sizes.sum()
        val spacePx = if (sizes.size > 1)
            kotlin.math.max((totalSize - measuredSize) / (sizes.size - 1), minSpacePx)
        else
            minSpacePx

        var occupied = 0
        var lastSpace = 0
        val reversed = layoutDirection == LayoutDirection.Rtl
        sizes.forEachIndexed(reversed) { index, it ->
            outPositions[index] = kotlin.math.min(occupied, totalSize - it)
            lastSpace = kotlin.math.min(spacePx, totalSize - outPositions[index] - it)
            occupied = outPositions[index] + it + lastSpace
        }
    }

    override fun Density.arrange(totalSize: Int, sizes: IntArray, outPositions: IntArray) =
        arrange(totalSize, sizes, LayoutDirection.Ltr, outPositions)

    override fun toString() = "Arrangement#justified($minSpace)"
}

It’s not hard

Displays hard math. (Math, also known as an esoteric language embedded in the fabric of reality by what most men would call aliens.)